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How to Solve Logarithms

Last Updated: October 5, 2023 Fact Checked

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 243,497 times.

Logarithms might be intimidating, but solving a logarithm is much simpler once you realize that logarithms are just another way to write out exponential equations. Once you rewrite the logarithm into a more familiar form, you should be able to solve it as you would solve any standard exponential equation.

Before You Begin: Learn to Express a Logarithmic Equation Exponentially [1] X Research source [2] X Research source

Step 1 Know the logarithm definition.

  • If and only if: b y = x
  • b does not equal 1
  • In the same equation, y is the exponent and x is the exponential expression that the logarithm is set equal to.

Step 2 Look at the equation.

  • Example: 1024 = ?

Step 4 Apply the exponent to the base.

  • This could also be written as: 4 5

Step 5 Rewrite your final answer.

  • Example: 4 5 = 1024

Method One: Solve for X

Step 1 Isolate the logarithm.

  • log 3 ( x + 5) + 6 - 6 = 10 - 6
  • log 3 ( x + 5) = 4

Step 2 Rewrite the equation in exponential form.

  • Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 4; b = 3; x = x + 5
  • Rewrite the equation so that: b y = x
  • 3 4 = x + 5

Step 3 Solve for x.

  • 3 * 3 * 3 * 3 = x + 5
  • 81 - 5 = x + 5 - 5

Step 4 Write your final answer.

  • Example: x = 76

Method Two: Solve for X Using the Logarithmic Product Rule [3] X Research source [4] X Research source

Step 1 Know the product rule.

  • log b (m * n) = log b (m) + log b (n)

Step 2 Isolate the logarithm to one side of the equation.

  • log 4 (x + 6) + log 4 (x) = 2 - log 4 (x) + log 4 (x)
  • log 4 (x + 6) + log 4 (x) = 2

Step 3 Apply the product rule.

  • log 4 [(x + 6) * x] = 2
  • log 4 (x 2 + 6x) = 2

Step 4 Rewrite the equation in exponential form.

  • Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 2; b = 4 ; x = x 2 + 6x
  • 4 2 = x 2 + 6x

Step 5 Solve for x.

  • 4 * 4 = x 2 + 6x
  • 16 = x 2 + 6x
  • 16 - 16 = x 2 + 6x - 16
  • 0 = x 2 + 6x - 16
  • 0 = (x - 2) * (x + 8)
  • x = 2; x = -8

Step 6 Write your answer.

  • Example: x = 2
  • Note that you cannot have a negative solution for a logarithm, so you can discard x - 8 as a solution.

Method Three: Solve for X Using the Logarithmic Quotient Rule [5] X Research source

Step 1 Know the quotient rule.

  • log b (m / n) = log b (m) - log b (n)

Step 2 Isolate the logarithm to one side of the equation.

  • log 3 (x + 6) - log 3 (x - 2) = 2 + log 3 (x - 2) - log 3 (x - 2)
  • log 3 (x + 6) - log 3 (x - 2) = 2

Step 3 Apply the quotient rule.

  • log 3 [(x + 6) / (x - 2)] = 2

Step 4 Rewrite the equation in exponential form.

  • Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 2; b = 3; x = (x + 6) / (x - 2)
  • 3 2 = (x + 6) / (x - 2)

Step 5 Solve for x.

  • 3 * 3 = (x + 6) / (x - 2)
  • 9 = (x + 6) / (x - 2)
  • 9 * (x - 2) = [(x + 6) / (x - 2)] * (x - 2)
  • 9x - 18 = x + 6
  • 9x - x - 18 + 18 = x - x + 6 + 18
  • 8x / 8 = 24 / 8

Step 6 Write your final answer.

  • Example: x = 3

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Community Q&A

Community Answer

  • ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut43_logfun.htm#logdef
  • ↑ https://www.mathsisfun.com/algebra/logarithms.html
  • ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut46_logeq.htm
  • ↑ https://www.youtube.com/watch?v=fnhFneOz6n8
  • ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut44_logprop.htm

About This Article

Grace Imson, MA

To solve a logarithm, start by identifying the base, which is "b" in the equation, the exponent, which is "y," and the exponential expression, which is "x." Then, move the exponential expression to one side of the equation, and apply the exponent to the base by multiplying the base by itself the number of times indicated in the exponent. Finally, rewrite your final answer as an exponential expression. To learn how to solve for "x" in a logarithm, scroll down! Did this summary help you? Yes No

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how to solve a problem with log

How To Solve Logarithmic Equations

Step By Step Video and practice Problems

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how to solve a problem with log

How To Solve Logarithmic Equations Video

What is the general strategy for solving log equations?

Answer: As the video above points out, there are two main types of logarithmic equations. Before you to decide how to solve an equation, you must determine whether the equation

  • A) has a logarithm on one side and a number on the other
  • B) whether it has logarithms on both sides

Example 1 Logarithm on one side and a number on the other

$$ log_4 x + log_4 8 = 3 $$

Step 1 Rewrite log side as single logarithm

$$ log_4 8x = 3 $$

Step 2 Rewrite as exponential equation

$$ 4^ 3 = 8x $$

Step 3 Solve the exponential equation

64 = 8x 8 = x

Example 2 Logarithm on both sides

Step 1 use the rules of logarithms to rewrite the left side and the right side of the equation to a single logarithm

Step 2 "cancel" the log

Step 3 solve the expression

Let's look at a specific ex $$ log_5 x + log_2 3 = log_5 6 $$

Step 1 rewrite both sides as single logs

$$ log_5 x + log_5 2 = log_5 6 \\ log_5 2x = log_5 6 $$

Step 2 "cancel" logs

$$ \color{Red}{ \cancel {log_5}} 2x = \color{Red}{ \cancel {log_5}} 6 \\ 2x = 6 $$

Step 2 Solve expression

Practice Problems

Solve the following equation: $ log_3 5 + log_3 x = log_3 15 $

Follow the steps for solving logarithmic equations with logs on both sides

rewrite both sides as single logs

$$ log_3 5x = log_3 15 $$

"cancel" logs

$$ \color{Red}{ \cancel{log_3}} 5x = \color{Red}{ \cancel{log_3}} 15 \\ 5x=15 $$

Solve expression

Solve the equation below: $ log_3 9 + log_3 x = 4 $

Follow the steps for solving logarithmic equations with a log on one side

Rewrite log side as single logarithm

$$ log_3 9x = 4$$

Rewrite as exponential equation

$$ 3^4 = 9x $$

Solve exponential equation

81 = 9 x 9 = x

Solve the following equation: $ 2log_3 5 + log_3 x = 3log_3 $

Follow the steps on how to solve equations with logs on both sides

rewrite both sides as single logs<

$ log_3 5^2 + log_3 x = log_3 5^3 \\ log_3 25 +log_3 x = log_3 125 \\ log_3 25x = log_3 125 $

$ \color{Red}{ \cancel{log_3}} 25x = \color{Red}{ \cancel{log_3}} 125 \\ 25x = 125 $

Solve the equation below: $ 2 log_2 4 + log_2 x = 5 $

$ 2 log_2 4 + log_2 x = 5 \\ log_2 4^2 = log_2 x = 5 \\ log_2 16 + log_2 x = 5 \\ log_2 16x = 5 $

32 =16x 2 = x

Solve the following equation: $ 2 log_3 5 + log_3 x = 3 log_3 5 $

$ log_3 5^2 + log_3 x + log_3 5^3 \\ log_3 25x + log_3 125 $

log 3 25x = log 3 5 3

$ \color{Red}{ \cancel{log_3}} 25x + \color{Red}{ \cancel{log_3}} 125 \\ 25x=125 $

$ \frac{25x}{25} = \frac{125}{25} \\ $

Solve the following equation: $2 log_3 7 - log_3 2x = log_3 98$

$ 2 log_3 7 - log_3 2x = log_3 98 \\ log_3 7^2 - log_3 2x = log_3 98 \\ log_3 49 - log_3 2x = log_3 98 \\ log_3 \frac{49}{2x} = log_3 98 $

$ \color{Red}{ \cancel{log_3}} \frac{49}{2x} = \color{Red}{ \cancel{log_3}} 98 \\ \frac{49}{2x} = 98 $

$ 49 = 196x \\ \frac{49}{196} = x \\ x = 49 $

Solve the following equation: $ 2 log_11 5 + log_11 x + log_11 2 = log_11 150 $

You know the deal. Just follow the steps for solving logarithmic equations with logs on both sides

rewrite as single logs

$ 2 log_11 5 + log_11 x + log_11 2 = log_11 150 \\ log_11 5^2 + log_11 2x = log_11 150 \\ log_11 25 + log_11 2x = log_11 150 \\ log_11 50x= log_11 150 $

2log 11 5 + log 11 x+ log 11 2 = log 11 150

$ \color{Red}{ \cancel{log_1}} 50x = \color{Red}{ \cancel{log_11}} 150 \\ 50x = 150 $

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Mathematics LibreTexts

4.7: Exponential and Logarithmic Equations

  • Last updated
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  • Page ID 64854

Learning Objectives

  • Use like bases to solve exponential equations.
  • Use logarithms to solve exponential equations.
  • Use the definition of a logarithm to solve logarithmic equations.
  • Use the one-to-one property of logarithms to solve logarithmic equations.
  • Solve applied problems involving exponential and logarithmic equations.

In 1859, an Australian landowner named Thomas Austin released \(24\) rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.

Seven rabbits in front of a brick building.

Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions.

Using Like Bases to Solve Exponential Equations

The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers \(b\), \(S\), and \(T\), where \(b>0\), \(b≠1\), \(b^S=b^T\) if and only if \(S=T\).

In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown.

For example, consider the equation \(3^{4x−7}=\dfrac{3^{2x}}{3}\). To solve for \(x\), we use the division property of exponents to rewrite the right side so that both sides have the common base, \(3\). Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for \(x\):

\[\begin{align*} 3^{4x-7}&= \dfrac{3^{2x}}{3}\\ 3^{4x-7}&= \dfrac{3^{2x}}{3^1} \qquad &&\text{Rewrite 3 as } 3^1\\ 3^{4x-7}&= 3^{2x-1} \qquad &&\text{Use the division property of exponents}\\ 4x-7&= 2x-1 \qquad &&\text{Apply the one-to-one property of exponents}\\ 2x&= 6 \qquad &&\text{Subtract 2x and add 7 to both sides}\\ x&= 3 \qquad &&\text{Divide by 3} \end{align*}\]

USING THE ONE-TO-ONE PROPERTY OF EXPONENTIAL FUNCTIONS TO SOLVE EXPONENTIAL EQUATIONS

For any algebraic expressions \(S\) and \(T\), and any positive real number \(b≠1\),

\[\begin{align} b^S=b^T\text{ if and only if } S=T \end{align}\]

How to: Given an exponential equation with the form \(b^S=b^T\), where \(S\) and \(T\) are algebraic expressions with an unknown, solve for the unknown.

  • Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form \(b^S=b^T\).
  • Use the one-to-one property to set the exponents equal.
  • Solve the resulting equation, \(S=T\), for the unknown.

Example \(\PageIndex{1}\): Solving an Exponential Equation with a Common Base

Solve \(2^{x−1}=2^{2x−4}\).

\[\begin{align*} 2^{x-1}&= 2^{2x-4} \qquad &&\text{The common base is 2}\\ x-1&= 2x-4 \qquad &&\text{By the one-to-one property the exponents must be equal}\\ x&= 3 \qquad &&\text{Solve for x} \end{align*}\]

Exercise \(\PageIndex{1}\)

Solve \(5^{2x}=5^{3x+2}\).

\(x=−2\)

Rewrite Equations So All Powers Have the Same Base

Sometimes the common base for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property.

For example, consider the equation \(256=4^{x−5}\). We can rewrite both sides of this equation as a power of \(2\). Then we apply the rules of exponents, along with the one-to-one property, to solve for \(x\):

\[\begin{align*} 256&= 4^{x-5}\\ 2^8&= {(2^2)}^{x-5} \qquad &&\text{Rewrite each side as a power with base 2}\\ 2^8&= 2^{2x-10} \qquad &&\text{Use the one-to-one property of exponents}\\ 8&= 2x-10 \qquad &&\text{Apply the one-to-one property of exponents}\\ 18&= 2x \qquad &&\text{Add 10 to both sides}\\ x&= 9 \qquad &&\text{Divide by 2} \end{align*}\]

How to: Given an exponential equation with unlike bases, use the one-to-one property to solve it.

  • Rewrite each side in the equation as a power with a common base.

Example \(\PageIndex{2}\): Solving Equations by Rewriting Them to Have a Common Base

Solve \(8^{x+2}={16}^{x+1}\).

\[\begin{align*} 8^{x+2}&= {16}^{x+1}\\ {(2^3)}^{x+2}&= {(2^4)}^{x+1} \qquad &&\text{Write 8 and 16 as powers of 2}\\ 2^{3x+6}&= 2^{4x+4} \qquad &&\text{To take a power of a power, multiply exponents}\\ 3x+6&= 4x+4 \qquad &&\text{Use the one-to-one property to set the exponents equal}\\ x&= 2 \qquad &&\text{Solve for x} \end{align*}\]

Exercise \(\PageIndex{2}\)

Solve \(5^{2x}={25}^{3x+2}\).

\(x=−1\)

Example \(\PageIndex{3}\): Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base

Solve \(2^{5x}=\sqrt{2}\).

\[\begin{align*} 2^{5x}&= 2^{\frac{1}{2}} \qquad &&\text{Write the square root of 2 as a power of 2}\\ 5x&= \dfrac{1}{2} \qquad &&\text{Use the one-to-one property}\\ x&= \dfrac{1}{10} \qquad &&\text{Solve for x} \end{align*}\]

Exercise \(\PageIndex{3}\)

Solve \(5^x=\sqrt{5}\).

\(x=\dfrac{1}{2}\)

Q&A: Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?

No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined.

Example \(\PageIndex{4}\): Solving an Equation with Positive and Negative Powers

Solve \(3^{x+1}=−2\).

This equation has no solution. There is no real value of \(x\) that will make the equation a true statement because any power of a positive number is positive.

Figure \(\PageIndex{2}\) shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.

Graph of 3^(x+1)=-2 and y=-2. The graph notes that they do not cross.

Exercise \(\PageIndex{4}\)

Solve \(2^x=−100\).

The equation has no solution.

Solving Exponential Equations Using Logarithms

Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since \(\log(a)=\log(b)\) is equivalent to \(a=b\), we may apply logarithms with the same base on both sides of an exponential equation.

How to: Given an exponential equation in which a common base cannot be found, solve for the unknown

  • If one of the terms in the equation has base 10, use the common logarithm.
  • If none of the terms in the equation has base 10, use the natural logarithm.
  • Use the rules of logarithms to solve for the unknown.

Example \(\PageIndex{5}\): Solving an Equation Containing Powers of Different Bases

Solve \(5^{x+2}=4^x\).

\[\begin{align*} 5^{x+2}&= 4^x \qquad &&\text{There is no easy way to get the powers to have the same base}\\ \ln5^{x+2}&= \ln4^x \qquad &&\text{Take ln of both sides}\\ (x+2)\ln5&= x\ln4 \qquad &&\text{Use laws of logs}\\ x\ln5+2\ln5&= x\ln4 \qquad &&\text{Use the distributive law}\\ x\ln5-x\ln4&= -2\ln5 \qquad &&\text{Get terms containing x on one side, terms without x on the other}\\ x(\ln5-\ln4)&= -2\ln5 \qquad &&\text{On the left hand side, factor out an x}\\ x\ln \left (\dfrac{5}{4} \right )&= \ln \left (\dfrac{1}{25} \right ) \qquad &&\text{Use the laws of logs}\\ x&=\dfrac{\ln \left (\dfrac{1}{25} \right )}{\ln \left (\dfrac{5}{4} \right )} \qquad &&\text{Divide by the coefficient of x} \end{align*}\]

Exercise \(\PageIndex{5}\)

Solve \(2^x=3^{x+1}\).

\(x=\dfrac{\ln3}{\ln \left (\dfrac{2}{3} \right )}\)

Q&A: Is there any way to solve \(2^x=3^x\)?

Yes. The solution is \(0\).

Equations Containing \(e\)

One common type of exponential equations are those with base \(e\). This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base \(e\) on either side, we can use the natural logarithm to solve it.

How to: Given an equation of the form \(y=Ae^{kt}\), solve for \(t\).

  • Divide both sides of the equation by \(A\).
  • Apply the natural logarithm of both sides of the equation.
  • Divide both sides of the equation by \(k\).

Example \(\PageIndex{6}\): Solve an Equation of the Form \(y = Ae^{kt}\)

Solve \(100=20e^{2t}\).

\[\begin{align*} 100&= 20e^{2t}\\ 5&= e^{2t} \qquad &&\text{Divide by the coefficient of the power}\\ \ln5&= 2t \qquad &&\text{Take ln of both sides. Use the fact that } ln(x) \text{ and } e^x \text{ are inverse functions}\\ t&= \dfrac{\ln5}{2} \qquad &&\text{Divide by the coefficient of t} \end{align*}\]

Using laws of logs, we can also write this answer in the form \(t=\ln\sqrt{5}\). If we want a decimal approximation of the answer, we use a calculator.

Exercise \(\PageIndex{6}\)

Solve \(3e^{0.5t}=11\).

\(t=2\ln \left (\dfrac{11}{3} \right )\) or \(\ln{ \left (\dfrac{11}{3} \right )}^2\)

Q&A: Does every equation of the form \(y=Ae^{kt}\) have a solution?

No. There is a solution when \(k≠0\),and when \(y\) and \(A\) are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is \(2=−3e^t\).

Example \(\PageIndex{7}\): Solving an Equation That Can Be Simplified to the Form \(y=Ae^{kt}\)

Solve \(4e^{2x}+5=12\).

\[\begin{align*} 4e^{2x}+5&= 12\\ 4e^{2x}&= 7 \qquad &&\text{Combine like terms}\\ e^{2x}&= \dfrac{7}{4} \qquad &&\text{Divide by the coefficient of the power}\\ 2x&= \ln \left (\dfrac{7}{4} \right ) \qquad &&\text{Take ln of both sides}\\ x&= \dfrac{1}{2}\ln \left (\dfrac{7}{4} \right ) \qquad &&\text{Solve for x} \end{align*}\]

Exercise \(\PageIndex{7}\)

Solve \(3+e^{2t}=7e^{2t}\).

\(t=\ln \left (\dfrac{1}{\sqrt{2}} \right )=−\dfrac{1}{2}\ln(2)\)

Extraneous Solutions

Sometimes the methods used to solve an equation introduce an extraneous solution , which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.

Example \(\PageIndex{8}\): Solving Exponential Functions in Quadratic Form

Solve \(e^{2x}−e^x=56\).

\[\begin{align*} e^{2x}-e^x&= 56\\ e^{2x}-e^x-56&= 0 \qquad &&\text{Get one side of the equation equal to zero}\\ (e^x+7)(e^x-8)&= 0 \qquad &&\text{Factor by the FOIL method}\\ e^x+7&= 0 \qquad &&\text{or} \\ e^x-8&= 0 \qquad &&\text{If a product is zero, then one factor must be zero}\\ e^x&= -7 \qquad &&\text{or} \\ e^x&= 8 \qquad &&\text{Isolate the exponentials}\\ e^x&= 8 \qquad &&\text{Reject the equation in which the power equals a negative number}\\ x&= \ln8 \qquad &&\text{Solve the equation in which the power equals a positive number} \end{align*}\]

When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation \(e^x=−7\) because a positive number never equals a negative number. The solution \(\ln(−7)\) is not a real number, and in the real number system this solution is rejected as an extraneous solution.

Exercise \(\PageIndex{8}\)

Solve \(e^{2x}=e^x+2\).

Q&A: Does every logarithmic equation have a solution?

No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.

Using the Definition of a Logarithm to Solve Logarithmic Equations

We have already seen that every logarithmic equation \({\log}_b(x)=y\) is equivalent to the exponential equation \(b^y=x\). We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.

For example, consider the equation \({\log}_2(2)+{\log}_2(3x−5)=3\). To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for \(x\):

\[\begin{align*} {\log}_2(2)+{\log}_2(3x-5)&= 3\\ {\log}_2(2(3x-5))&= 3 \qquad \text{Apply the product rule of logarithms}\\ {\log}_2(6x-10)&= 3 \qquad \text{Distribute}\\ 2^3&= 6x-10 \qquad \text{Apply the definition of a logarithm}\\ 8&= 6x-10 \qquad \text{Calculate } 2^3\\ 18&= 6x \qquad \text{Add 10 to both sides}\\ x&= 3 \qquad \text{Divide by 6} \end{align*}\]

USING THE DEFINITION OF A LOGARITHM TO SOLVE LOGARITHMIC EQUATIONS

For any algebraic expression \(S\) and real numbers \(b\) and \(c\),where \(b>0\), \(b≠1\),

\[\begin{align} {\log}_b(S)=c \text{ if and only if } b^c=S \end{align}\]

Example \(\PageIndex{9}\): Using Algebra to Solve a Logarithmic Equation

Solve \(2\ln x+3=7\).

\[\begin{align*} 2\ln x+3&= 7\\ 2\ln x&= 4 \qquad \text{Subtract 3}\\ \ln x&= 2 \qquad \text{Divide by 2}\\ x&= e^2 \qquad \text{Rewrite in exponential form} \end{align*}\]

Exercise \(\PageIndex{9}\)

Solve \(6+\ln x=10\).

Example \(\PageIndex{10}\): Using Algebra Before and After Using the Definition of the Natural Logarithm

Solve \(2\ln(6x)=7\).

\[\begin{align*} 2\ln(6x)&= 7\\ \ln(6x)&= \dfrac{7}{2} \qquad \text{Divide by 2}\\ 6x&= e^{\left (\dfrac{7}{2} \right )} \qquad \text{Use the definition of }\ln \\ x&= \dfrac{1}{6}e^{\left (\dfrac{7}{2} \right )} \qquad \text{Divide by 6} \end{align*}\]

Exercise \(\PageIndex{10}\)

Solve \(2\ln(x+1)=10\).

\(x=e^5−1\)

Example \(\PageIndex{11}\): Using a Graph to Understand the Solution to a Logarithmic Equation

Solve \(\ln x=3\).

\[\begin{align*} \ln x&= 3\\ x&= e^3 \qquad \text{Use the definition of the natural logarithm} \end{align*}\]

Figure \(\PageIndex{3}\) represents the graph of the equation. On the graph, the x -coordinate of the point at which the two graphs intersect is close to \(20\). In other words \(e^3≈20\). A calculator gives a better approximation: \(e^3≈20.0855\).

Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).

Exercise \(\PageIndex{11}\)

Use a graphing calculator to estimate the approximate solution to the logarithmic equation \(2^x=1000\) to \(2\) decimal places.

\(x≈9.97\)

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers \(x>0\), \(S>0\), \(T>0\) and any positive real number \(b\), where \(b≠1\),

\({\log}_bS={\log}_bT\) if and only if \(S=T\).

For example,

If \({\log}_2(x−1)={\log}_2(8)\), then \(x−1=8\).

So, if \(x−1=8\), then we can solve for \(x\),and we get \(x=9\). To check, we can substitute \(x=9\) into the original equation: \({\log}_2(9−1)={\log}_2(8)=3\). In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.

For example, consider the equation \(\log(3x−2)−\log(2)=\log(x+4)\). To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for \(x\):

\[\begin{align*} \log(3x-2)-\log(2)&= \log(x+4)\\ \log \left (\dfrac{3x-2}{2} \right )&= \log(x+4) \qquad \text{Apply the quotient rule of logarithms}\\ \dfrac{3x-2}{2}&= x+4 \qquad \text{Apply the one to one property of a logarithm}\\ 3x-2&= 2x+8 \qquad \text{Multiply both sides of the equation by 2}\\ x&= 10 \qquad \text{Subtract 2x and add 2} \end{align*}\]

To check the result, substitute \(x=10\) into \(\log(3x−2)−\log(2)=\log(x+4)\).

\[\begin{align*} \log(3(10)-2)-\log(2)&= \log((10)+4) \\ \log(28)-\log(2)&= \log(14)\\ \log \left (\dfrac{28}{2} \right )&= \log(14) \qquad \text{The solution checks} \end{align*}\]

USING THE ONE-TO-ONE PROPERTY OF LOGARITHMS TO SOLVE LOGARITHMIC EQUATIONS

For any algebraic expressions \(S\) and \(T\) and any positive real number \(b\), where \(b≠1\),

Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.

How to: Given an equation containing logarithms, solve it using the one-to-one property

  • Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form \({\log}_bS={\log}_bT\).
  • Use the one-to-one property to set the arguments equal.

Example \(\PageIndex{12}\): Solving an Equation Using the One-to-One Property of Logarithms

Solve \(\ln(x^2)=\ln(2x+3)\).

\[\begin{align*} \ln(x^2)&= \ln(2x+3)\\ x^2&= 2x+3 \qquad \text{Use the one-to-one property of the logarithm}\\ x^2-2x-3&= 0 \qquad \text{Get zero on one side before factoring}\\ (x-3)(x+1)&= 0 \qquad \text{Factor using FOIL}\\ x-3&= 0 \qquad \text{or } x+1=0 \text{ If a product is zero, one of the factors must be zero}\\ x=3 \qquad \text{or} \\ x&= -11 \qquad \text{Solve for x} \end{align*}\]

There are two solutions: \(3\) or \(−1\). The solution \(−1\) is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.

Exercise \(\PageIndex{12}\)

Solve \(\ln(x^2)=\ln1\).

\(x=1\) or \(x=−1\)

Solving Applied Problems Using Exponential and Logarithmic Equations

In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.

One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life . Table \(\PageIndex{1}\) lists the half-life for several of the more common radioactive substances.

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:

\[\begin{align} A(t)&= A_0e^{\tfrac{\ln(0.5)}{T}t}\\ A(t)&= A_0e^{\tfrac{\ln(0.5)t}{T}}\\ A(t)&= A_0{(e^{\ln(0.5)})}^{\tfrac{t}{T}}\\ A(t)&= A_0{\left (\dfrac{1}{2}\right )}^{\tfrac{t}{T}}\\ \end{align}\]

  • \(A_0\) is the amount initially present
  • \(T\) is the half-life of the substance
  • \(t\) is the time period over which the substance is studied
  • \(y\) is the amount of the substance present after time \(t\)

Example \(\PageIndex{13}\): Using the Formula for Radioactive Decay to Find the Quantity of a Substance

How long will it take for ten percent of a \(1000\)-gram sample of uranium-235 to decay?

\[\begin{align*} y&= 1000e^{\tfrac{\ln(0.5)}{703,800,000}t}\\ 900&= 1000e^{\tfrac{\ln(0.5)}{703,800,000}t} \qquad \text{After } 10\% \text{ decays, 900 grams are left}\\ 0.9&= e^{\tfrac{\ln(0.5)}{703,800,000}t} \qquad \text{Divide by 1000}\\ \ln(0.9)&= \ln \left (e^{\tfrac{\ln(0.5)}{703,800,000}t} \right ) \qquad \text{Take ln of both sides}\\ \ln(0.9)&= \dfrac{\ln(0.5)}{703,800,000}t \qquad \ln(e^M)=M\\ t&= 703,800,000\times \dfrac{\ln(0.9)}{\ln(0.5)} \qquad \text{years Solve for t}\\ t&\approx 106,979,777 \qquad \text{years} \end {align*} \]

Ten percent of \(1000\) grams is \(100\) grams. If \(100\) grams decay, the amount of uranium-235 remaining is \(900\) grams.

Exercise \(\PageIndex{13}\)

How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?

\(t=703,800,000×\dfrac{\ln(0.8)}{\ln(0.5)}\)years ≈ 226,572,993 years.

Access these online resources for additional instruction and practice with exponential and logarithmic equations.

  • Solving Logarithmic Equations
  • Solving Exponential Equations with Logarithms

Key Equations

Key concepts.

  • We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.
  • When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown. See Example \(\PageIndex{1}\).
  • When we are given an exponential equation where the bases are not explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown. See Example \(\PageIndex{2}\), Example \(\PageIndex{3}\), and Example \(\PageIndex{4}\).
  • When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side. See Example \(\PageIndex{5}\).
  • We can solve exponential equations with base \(e\),by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other. See Example \(\PageIndex{6}\) and Example \(\PageIndex{7}\).
  • After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions. See Example \(\PageIndex{8}\).
  • When given an equation of the form \({\log}_b(S)=c\), where \(S\) is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation \(b^c=S\), and solve for the unknown. See Example \(\PageIndex{9}\) and Example \(\PageIndex{10}\).
  • We can also use graphing to solve equations with the form \({\log}_b(S)=c\). We graph both equations \(y={\log}_b(S)\) and \(y=c\) on the same coordinate plane and identify the solution as the x- value of the intersecting point. See Example \(\PageIndex{11}\).
  • When given an equation of the form \({\log}_bS={\log}_bT\), where \(S\) and \(T\) are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation \(S=T\) for the unknown. See Example \(\PageIndex{12}\).
  • Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm. See Example \(\PageIndex{13}\).

Introduction to Logarithms

In its simplest form, a logarithm answers the question:

How many of one number multiply together to make another number?

Example: How many 2 s multiply together to make 8 ?

Answer: 2 × 2 × 2 = 8 , so we had to multiply 3 of the 2 s to get 8

So the logarithm is 3

How to Write it

We write it like this:

log 2 (8) = 3

 So these two things are the same:

The number we multiply is called the "base", so we can say:

  • "the logarithm of 8 with base 2 is 3"
  • or "log base 2 of 8 is 3"
  • or "the base-2 log of 8 is 3"

Notice we are dealing with three numbers:

  • the base : the number we are multiplying (a "2" in the example above)
  • how often to use it in a multiplication (3 times, which is the logarithm )
  • The number we want to get (an "8")

More Examples

Example: what is log 5 (625) ... .

We are asking "how many 5s need to be multiplied together to get 625?"

5 × 5 × 5 × 5 = 625 , so we need 4 of the 5s

Answer: log 5 (625) = 4

Example: What is log 2 (64) ... ?

We are asking "how many 2s need to be multiplied together to get 64?"

2 × 2 × 2 × 2 × 2 × 2 = 64 , so we need 6 of the 2s

Answer: log 2 (64) = 6

Exponents and Logarithms are related, let's find out how ...

So a logarithm answers a question like this:

In this way:

The logarithm tells us what the exponent is!

In that example the "base" is 2 and the "exponent" is 3:

So the logarithm answers the question:

What exponent do we need (for one number to become another number) ?

The general case is:

Example: What is log 10 (100) ... ?

So an exponent of 2 is needed to make 10 into 100, and:

log 10 (100) = 2

Example: What is log 3 (81) ... ?

So an exponent of 4 is needed to make 3 into 81, and:

log 3 (81) = 4

Common Logarithms: Base 10

Sometimes a logarithm is written without a base, like this:

This usually means that the base is really 10 .

It is called a "common logarithm". Engineers love to use it.

On a calculator it is the "log" button.

It is how many times we need to use 10 in a multiplication, to get our desired number.

Example: log(1000) = log 10 (1000) = 3

Natural Logarithms: Base "e"

Another base that is often used is e (Euler's Number) which is about 2.71828.

This is called a "natural logarithm". Mathematicians use this one a lot.

On a calculator it is the "ln" button.

It is how many times we need to use "e" in a multiplication, to get our desired number.

Example: ln(7.389) = log e ( 7.389 ) ≈ 2

Because 2.71828 2 ≈ 7.389

But Sometimes There Is Confusion ... !

Mathematicians may use "log" (instead of "ln") to mean the natural logarithm. This can lead to confusion:

So, be careful when you read "log" that you know what base they mean!

Logarithms Can Have Decimals

All of our examples have used whole number logarithms (like 2 or 3), but logarithms can have decimal values like 2.5, or 6.081, etc.

Example: what is log 10 (26) ... ?

The logarithm is saying that 10 1.41497... = 26 (10 with an exponent of 1.41497... equals 26)

Read Logarithms Can Have Decimals to find out more.

Negative Logarithms

A negative logarithm means how many times to divide by the number.

We can have just one divide:

Example: What is log 8 (0.125) ... ?

Well, 1 ÷ 8 = 0.125 ,

So log 8 (0.125) = −1

Or many divides:

Example: What is log 5 (0.008) ... ?

1 ÷ 5 ÷ 5 ÷ 5 = 5 -3 ,

So log 5 (0.008) = −3

It All Makes Sense

Multiplying and Dividing are all part of the same simple pattern.

Let us look at some Base-10 logarithms as an example:

Looking at that table, see how positive, zero or negative logarithms are really part of the same (fairly simple) pattern.

"Logarithm" is a word made up by Scottish mathematician John Napier (1550-1617), from the Greek word logos meaning "proportion, ratio or word" and arithmos meaning "number", ... which together makes "ratio-number" !

how to solve a problem with log

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Logarithmic Word Problems

Log Probs Expo Growth Expo Decay

What are logarithm word problems?

Logarithmic word problems, in my experience, generally involve either evaluating a given logarithmic equation at a given point, or else solving an equation for a given variable; they're pretty straightforward.

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What real-world problems use logarithms?

The classic real-world contexts for logarithm word problems are the measurement of acidity or alkalinity (that is, the measurement of pH), the measurement of sound (in decibels, or dB), and the measurement of earthquake intensity (on the Richter scale), among other uses ( link ).

Note: While log-based word problems are, in my experience, pretty straightforward, their statements tend to be fairly lengthy. Expect to have to plow through an unusual amount of text before they get to the point.

  • Chemists define the acidity or alkalinity of a substance according to the formula pH =  −log[H + ] where [H + ] is the hydrogen ion concentration, measured in moles per liter. Solutions with a pH value of less than 7 are acidic; solutions with a pH value of greater than 7 are basic; solutions with a pH of 7 (such as pure water) are neutral.

a) Suppose that you test apple juice and find that the hydrogen ion concentration is [H + ] = 0.0003 . Find the pH value and determine whether the juice is basic or acidic.

b) You test some ammonia and determine the hydrogen ion concentration to be [H + ] = 1.3 × 10 −9 . Find the pH value and determine whether the ammonia is basic or acidic.

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In each case, I need to evaluate the pH function at the given value of [H + ] . In other words, this exercise, despite all the verbiage, is just plug-n-chug.

Since no base is specified, I will assume that the base for this logarithm is 10 , so that this is the so-called "common" log. (I happen to know that 10 is indeed the correct base, but they should have specified.)

a) In the case of the apple juice, the hydrogen ion concentration is [H + ] = 0.0003 , so:

pH = −log[H + ]

= −log[0.0003]

= 3.52287874528...

This value is less than 7 , so the apple juice is acidic.

b) In the case of the ammonia, the hydrogen ion concentration is [H + ] = 1.3 × 10 −9 , so:

= −log[1.3 × 10 −9 ] = 8.88605664769...

This value is more than 7 , so the ammonia is basic.

(a) The juice is acidic with a pH of about 3.5 , and (b) the ammonia is basic with a pH of about 8.9 .

When a logarithm is given without a base being specified, different people in different contexts will assume different bases; either 10 , 2 , or e . Ask now whether or not bases will be specified for all exercises, or if you're going to be expected to "just know" the bases for certain formulas, or if you're supposed to "just assume" that all logs without a specified base have a base of... [find out which one].

  • "Loudness" is measured in decibels (abbreviated as dB). The formula for the loudness of a sound is given by dB = 10×log[I ÷ I 0 ] where I 0 is the intensity of "threshold sound", or sound that can barely be perceived. Other sounds are defined in terms of how many times more intense they are than threshold sound. For instance, a cat's purr is about 316 times as intense as threshold sound, for a decibel rating of:

dB = 10×log[I ÷ I 0 ]     = 10×log[ (316 I 0 ) ÷ I 0 ]     = 10×log[ 316 ]     = 24.9968708262...

...about 25 decibels.

Considering that prolonged exposure to sounds above 85 decibels can cause hearing damage or loss, and considering that a gunshot from a .22 rimfire rifle has an intensity of about I = (2.5 × 10 13 )I 0 , should you follow the rules and wear ear protection when practicing at the rifle range?

I need to evaluate the decibel equation at I = (2.5 × 10 13 )I 0 :

dB = 10log [ I ÷ I 0 ]     = 10log[ (2.5 ×10 13 )I 0 ÷ I 0 ]     = 10log[2.5 ×10 13 ]     = 133.979400087...

In other words, the squirrel gun creates a noise level of about 134 decibels. Since this is well above the level at which I can suffer hearing damage,

I should follow the rules and wear ear protection.

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  • Earthquake intensity is measured by the Richter scale. The formula for the Richter rating of a given quake is given by R = log[ I ÷ I 0  ] where I 0 is the "threshold quake", or movement that can barely be detected, and the intensity I is given in terms of multiples of that threshold intensity.

You have a seismograph set up at home, and see that there was an event while you were out that had an intensity of I = 989I 0 . Given that a heavy truck rumbling by can cause a microquake with a Richter rating of 3 or 3.5 , and "moderate" quakes have a Richter rating of 4 or more, what was likely the event that occurred while you were out?

To determine the probable event, I need to convert the intensity of the mystery quake into a Richter rating by evaluating the Richter function at I = 989I 0 :

R = log[ I ÷ I 0 ]     = log[ 989I 0 ÷ I 0 ]     = log[989]     = 2.9951962916...

A Richter rating of about 3 is not high enough to have been a moderate quake.

The event was probably just a big truck.

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how to solve a problem with log

Inspiration & Information for Self-Improvement

How to Solve a Logarithmic Equation (Guide)

Welcome to our comprehensive guide on how to solve a logarithmic equation ! Whether you’re a math enthusiast or simply looking for a step-by-step solution, this article will provide you with the necessary tools to tackle logarithmic equations with confidence. In this guide, we’ll explore the properties and rules of logarithms, discuss common mistakes to avoid, and provide strategies to help you solve these equations efficiently. So let’s dive in and demystify the world of logarithmic equation solving!

how to solve a logarithmic equation

Key Takeaways:

  • Logarithmic equations can be solved using the one-to-one property of logarithms.
  • Isolate the logarithmic expression and equate the arguments to solve the equation.
  • Understanding the properties and rules of logarithmic equations is crucial for successful solving.
  • Check your solutions to ensure they satisfy the original logarithmic equation.
  • Practice, practice, practice! Familiarity with logarithmic equations will improve your problem-solving skills.

Table of Contents

Steps to Solve a Logarithmic Equation

When it comes to solving a logarithmic equation, there are specific steps you can follow to arrive at the solution. By understanding and implementing these steps, you can tackle logarithmic equations with confidence. Let’s explore the step-by-step process below:

  • Step 1: Use the properties of logarithms – Begin by isolating the logarithmic expression on one side of the equation. Apply the rules of logarithms, such as the product rule, quotient rule, and power rule, to simplify the equation.
  • Step 2: Rewrite as an exponential equation – Apply the definition of logarithms to rewrite the equation as an exponential equation. This step helps to eliminate the logarithm and express the equation in a different form.
  • Step 3: Solve the equation – Solve the resulting exponential equation using techniques such as factoring, applying the quadratic formula if necessary, or simplifying the equation further.
  • Step 4: Check your answer – Once you have obtained a solution, it is crucial to check whether it satisfies the original logarithmic equation. Substitute the found value back into the equation and verify if both sides are equal.

By following these steps, you can effectively solve logarithmic equations and find the values that satisfy the given equation. Let’s apply these steps to a few examples to further illustrate the process.

Solve the equation log 2 (x) = 4.

To solve this equation, we can use the one-to-one property of logarithms. By rewriting the equation as 2 4 = x, we find that x = 16.

Solve the equation log 5 (x + 3) = 2.

To solve this equation, we apply the definition of logarithms. By converting the equation to an exponential form, we have 5 2 = x + 3. Simplifying further, we find that x = 22.

These examples demonstrate how the steps mentioned above can be used to solve logarithmic equations effectively. Applying the properties of logarithms, rewriting as an exponential equation, solving, and checking the answers are key components of the process. With practice and familiarity, you can become proficient in solving various logarithmic equations.

Understanding Logarithmic Equations

Logarithmic equations are a fundamental concept in mathematics , particularly in higher-level math, physics, computer science, and engineering. They provide an alternative way to represent exponential equations and offer unique insights into various mathematical problems. To effectively solve logarithmic equations, it is crucial to have a solid understanding of their properties and rules.

One of the key properties of logarithmic equations is the one-to-one property. This property states that if you have an equation with logarithms of the same base on both sides, you can equate their arguments. By applying this property, you can simplify and manipulate logarithmic equations to isolate the variable and find its value.

Another important aspect of logarithmic equations is the ability to rewrite them as exponential equations . This transformation allows you to switch between logarithmic and exponential notation, making it easier to solve equations by applying familiar algebraic techniques.

To better grasp the rules and properties of logarithmic equations, let’s explore a table that summarizes some of the key concepts:

By familiarizing yourself with these rules and properties, you can confidently tackle logarithmic equations and apply the appropriate techniques to solve them.

What are Logarithmic Equations?

Logarithmic equations are a type of equation that involves logarithms. Logarithms are the inverse functions of exponents and allow us to solve for the exponent when the resulting value is known. Logarithms are written in the form logb(x) = y, where b is the base, x is the argument, and y is the exponent. Solving logarithmic equations involves applying the properties and rules of logarithms to isolate the variable and find its value.

Logarithmic equations play a crucial role in various fields such as mathematics, physics, computer science, and engineering. They provide a powerful tool for solving exponential equations in a different form. By using logarithms, we can simplify complex calculations and solve problems that involve exponential growth or decay.

To better understand the concept of logarithmic equations, let’s consider an example. Suppose we have the equation log2(x) = 3. This equation states that 2 raised to the power of 3 is equal to x. By evaluating 2^3, we find that x = 8. Therefore, the solution to the logarithmic equation log2(x) = 3 is x = 8.

In summary, logarithmic equations involve logarithms and provide a powerful tool for solving exponential equations. By understanding the properties and rules of logarithms, we can effectively isolate variables and find their values. Logarithmic equations are widely used in various disciplines and offer a valuable approach to problem-solving.

Logarithmic Notation and Exponential Equations

Logarithmic notation is a powerful tool for representing exponential equations in a different form. With logarithmic notation , we can express an equation as log b (x) = y, where b is the base, x is the argument, and y is the exponent. This notation allows us to solve for the exponent when the base and argument are known.

By understanding the relationship between logarithmic notation and exponential equations, we can easily transition between the two forms. For example, if we have the exponential equation 2 3 = 8, we can rewrite it using logarithmic notation as log 2 (8) = 3. This notation gives us a concise way to express the relationship between the base, argument, and exponent.

Logarithmic notation is especially useful when working with complex equations involving exponents. It simplifies the representation of exponential equations and provides a clear framework for solving them. By mastering logarithmic notation, you can tackle a wide range of mathematical problems with confidence.

Logarithmic Equations on Standardized Tests

Logarithmic equations are occasionally tested on standardized exams such as the ACT and SAT. These tests assess your understanding of logarithmic properties and your ability to apply them to solve equations. It is important to be familiar with the basic rules and properties of logarithms to successfully solve these types of equations.

On the ACT, basic logarithmic equations may appear at least once per test, while the SAT may include logarithmic equations in the Math section. These questions are designed to evaluate your problem-solving skills and your knowledge of logarithmic concepts. Understanding logarithmic equations is crucial for achieving a high score on these exams.

When encountering logarithmic equations on standardized tests , it is essential to carefully read and understand the problem before attempting to solve it. Pay attention to specific instructions and any given restrictions, such as the domain of the variable. Take your time to apply the appropriate logarithmic properties and rules to simplify the equation and find the correct solution.

Preparing for standardized tests involves practicing with various logarithmic equation problems to reinforce your understanding and improve your problem-solving abilities. By familiarizing yourself with different types of logarithmic equations, you can develop strategies and techniques to efficiently solve these problems within the time constraints of the exam. Remember to review the necessary logarithmic rules and properties before test day to ensure you are well-prepared.

Strategies to Solve Logarithmic Equations

To effectively solve logarithmic equations, it is important to employ various strategies that can simplify the process and enhance accuracy. Here are some helpful tips and techniques:

  • Isolate the logarithmic expression: Begin by isolating the logarithmic expression on one side of the equation. This can be done by applying the properties and rules of logarithms, such as combining logarithms with the same base or using the inverse property.
  • Rewrite the equation in exponential form: Once the logarithmic expression is isolated, rewrite the equation in exponential form. This allows you to convert the logarithmic equation into a more familiar algebraic equation, making it easier to solve.
  • Apply the rules and properties of logarithms: Utilize the rules and properties of logarithms to simplify the equation further. These rules include the product rule, quotient rule, and power rule, which can help simplify complex logarithmic expressions.
  • Check your answers for validity: After obtaining a solution, it is crucial to check whether the solution satisfies the original logarithmic equation. This step helps identify any potential extraneous solutions and ensures the accuracy of the final answer.

Additionally, for more challenging logarithmic equations, you can employ other strategies such as using calculators or implementing a “guess and check” method. Calculators can quickly solve complex logarithmic equations, while the “guess and check” method involves making educated guesses and verifying the validity of the solution. Practice and familiarity with logarithmic equations will enhance your problem-solving skills and enable you to tackle a wide range of logarithmic equations confidently.

Remember, solving logarithmic equations requires a solid understanding of the properties and rules of logarithms. By utilizing these strategies and practicing regularly, you can improve your ability to solve logarithmic equations and effectively apply them to various mathematical problems.

“By isolating the logarithmic expression, rewriting the equation in exponential form, and applying the properties of logarithms, I was able to solve the complex logarithmic equation and obtain the correct solution. This strategy helped me understand the relationship between logarithmic notation and exponential equations, and enhanced my problem-solving skills in logarithmic equations.” – Math enthusiast

Strategies to solve logarithmic equations involve isolating the logarithmic expression, rewriting the equation in exponential form, applying the rules and properties of logarithms, and checking the solutions for validity. These strategies simplify the solving process and enhance accuracy. Additionally, using calculators or implementing a “guess and check” method can be helpful for more challenging logarithmic equations. With practice and familiarity, you can improve your ability to solve logarithmic equations and apply them to various mathematical problems.

Examples of Logarithmic Equation Solving

To illustrate the process of solving logarithmic equations, let’s consider a few examples:

  • Solve the equation log2(x) = 4. (Solution: Using the one-to-one property, we can rewrite the equation as 2^4 = x, so x = 16.)
  • Solve the equation log5(x + 3) = 2. (Solution: By applying the definition of logarithm, we have 5^2 = x + 3, which simplifies to x = 22.)

These examples demonstrate how logarithmic equations can be solved using the properties and rules of logarithms. By following the steps outlined earlier, you can apply similar techniques to solve different logarithmic equations. Remember to isolate the logarithmic expressions, rewrite them in exponential form if needed, and solve for the variable. Checking your answers is crucial to ensure their validity and accuracy.

Exploring more examples and practicing regularly will help you gain confidence in solving logarithmic equations and improve your problem-solving skills. Pay attention to common mistakes, such as misapplying the rules of logarithms or forgetting to check for extraneous solutions. By familiarizing yourself with the principles behind logarithmic equations and utilizing helpful strategies, you’ll be well-equipped to tackle these types of problems.

Common Mistakes in Solving Logarithmic Equations

When it comes to solving logarithmic equations, there are a few common mistakes that many students make. These errors can lead to incorrect solutions or even confusion during the problem-solving process. By being aware of these common mistakes, you can avoid them and improve your accuracy in solving logarithmic equations.

One common mistake is forgetting to check for extraneous solutions. Logarithmic equations can sometimes yield solutions that do not satisfy the original equation. It’s important to check your answers by substituting them back into the original equation and ensuring that they make sense.

Another mistake is not applying the one-to-one property correctly. The one-to-one property of logarithms states that if log b (x) = log b (y), then x = y. This property allows us to equate the arguments of logarithms with the same base. Make sure to properly apply this property when solving logarithmic equations with multiple logarithms.

Lastly, misapplying the rules and properties of logarithms can lead to mistakes. Logarithms have specific rules and properties, such as the power rule, product rule, and quotient rule, which must be used correctly when simplifying or manipulating logarithmic expressions. Be cautious when applying these rules and double-check your steps to ensure their accuracy.

By being aware of these common mistakes and practicing regularly, you can improve your accuracy in solving logarithmic equations. Remember to carefully check your work, apply the one-to-one property correctly, and use the rules and properties of logarithms accurately. With practice and attention to detail, you can confidently solve logarithmic equations and enhance your overall math skills.

Logarithmic Equation Solvers and Calculators

When it comes to solving logarithmic equations, there are several tools available online that can make the process faster and more convenient. Logarithmic equation solvers and calculators are designed to handle complex calculations and provide accurate solutions with just a few clicks. These tools can be especially helpful for those who are new to logarithmic equations or need to quickly verify their manual calculations.

Using a logarithmic equation solver or calculator is simple. You input the given equation, select the appropriate options, and the tool will generate the solution for you. The results are typically displayed in a clear and easy-to-understand format, making it easy to interpret and apply them to your own problem-solving process. However, it’s important to note that relying solely on these tools without understanding the principles behind logarithmic equations can limit your mathematical growth.

Remember: While logarithmic equation solvers and calculators can be extremely useful, it’s still crucial to develop a solid understanding of the properties and rules of logarithms. This knowledge will help you verify the accuracy of the results obtained from these tools and enhance your overall problem-solving abilities.

Overall, logarithmic equation solvers and calculators can be valuable resources for both beginners and experienced mathematicians. They offer speed and efficiency in solving complex logarithmic equations, but should be used as tools to complement your understanding rather than replacements for manual calculations. By combining the benefits of these tools with a solid foundation in logarithmic equations, you can enhance your problem-solving skills and tackle even the most challenging logarithmic problems.

In conclusion, solving logarithmic equations requires a solid understanding of the properties and rules of logarithms. By following the step-by-step process outlined in this guide, you can confidently solve logarithmic equations and enhance your mathematical skills.

Remember to isolate the logarithmic expression, apply the appropriate rules and properties, and check your answers for validity. Avoid common mistakes such as misapplying the one-to-one property and forgetting to check for extraneous solutions.

Additionally, it is beneficial to practice regularly and familiarize yourself with various logarithmic equation examples . The more you engage with logarithmic equations, the more comfortable you will become in solving them.

While logarithmic equation solvers and calculators can be helpful tools, it is still crucial to understand the principles behind solving logarithmic equations manually. This will allow you to verify the accuracy of the results obtained and develop a deeper comprehension of logarithmic equations.

How do you solve a logarithmic equation?

To solve a logarithmic equation, you can use the one-to-one property for logarithms. If you are given an equation with a logarithm of the same base on both sides, you can equate the arguments. The steps to solve a logarithmic equation are as follows: 1) Use the rules of exponents to isolate the logarithmic expression on both sides of the equation. 2) Set the arguments equal to each other. 3) Solve the resulting equation. 4) Check your answers to ensure they satisfy the original logarithmic equation.

How do you solve a logarithmic equation with a logarithm on one side and a constant on the other side?

When solving a logarithmic equation with a logarithm on one side and a constant on the other side, you can follow these steps: 1) Use the properties of logarithms to isolate the log on one side. 2) Apply the definition of logarithm and rewrite it as an exponential equation. 3) Solve the resulting equation. 4) Check your answers to verify their validity.

What are logarithmic equations and why are they important?

Logarithmic equations are a way to represent exponential equations in a different form. Logarithms can be used to solve a variety of mathematical problems and are commonly encountered in higher-level math, physics, computer science, and engineering. It is important to understand the properties and rules of logarithmic equations, such as the one-to-one property and the ability to rewrite logarithmic equations as exponential equations.

What is logarithmic notation and how does it relate to exponential equations?

Logarithmic notation is a way to express exponential equations in a different form. In logarithmic notation, a logarithm is written as logb(x) = y, where b is the base, x is the argument, and y is the exponent. Logarithmic notation allows us to solve for the exponent when the base and argument are known. By understanding the relationship between logarithmic notation and exponential equations, we can easily transition between the two forms.

Do logarithmic equations appear on standardized tests?

Yes, logarithmic equations are occasionally tested on standardized exams such as the ACT and SAT. It is important to be familiar with the basic rules and properties of logarithms to solve these types of equations. Understanding logarithmic equations is crucial for success on these exams.

What strategies can I use to solve logarithmic equations?

To solve logarithmic equations, you can employ various strategies. Some helpful tips include isolating the logarithmic expression, rewriting the equation in exponential form, applying the rules and properties of logarithms, and checking your answers for validity. Practice and familiarity with logarithmic equations will improve your problem-solving skills.

Can you provide examples of solving logarithmic equations?

Sure! Here are a few examples of solving logarithmic equations : 1) Solve the equation log2(x) = 4. Solution: Using the one-to-one property, we can rewrite the equation as 2^4 = x, so x = 16. 2) Solve the equation log5(x + 3) = 2. Solution: By applying the definition of logarithm, we have 5^2 = x + 3, which simplifies to x = 22.

What are some common mistakes to avoid when solving logarithmic equations?

Common mistakes in solving logarithmic equations include forgetting to check for extraneous solutions, not applying the one-to-one property correctly, and misapplying the rules and properties of logarithms. Paying attention to detail, double-checking your steps, and practicing regularly can help minimize these errors and improve your accuracy in solving logarithmic equations.

Are there logarithmic equation solvers and calculators available?

Yes, there are various logarithmic equation solvers and calculators available online. These tools can help you quickly find the solutions to logarithmic equations by inputting the given equation and obtaining the solution. However, it is still important to understand the principles behind solving logarithmic equations manually to verify the accuracy of the results obtained from these solvers.

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Properties of logarithmic functions

Comparison of exponential function and logarithmic function, practice questions, solving logarithmic functions – explanation & examples.

Solving Log Function Title

Logarithms and exponents are two topics in mathematics that are closely related. Therefore it is useful we take a brief review of exponents.

An exponent is a form of writing the repeated multiplication of a number by itself. An exponential function is of the form f (x) = b y , where b > 0 < x and b ≠ 1. The quantity x is the number, b is the base, and y is the exponent or power.

For example , 32 = 2 × 2 × 2 × 2 × 2 = 2 2 .

Solving Log Function Exp and Log

On the other hand, the logarithmic function is defined as the inverse function of exponentiation. Consider again the exponential function f(x) = b y , where b > 0 < x and b ≠ 1. We can represent this function in logarithmic form as:

y = log b x

Then the logarithmic function is given by;

f(x) = log b x = y, where b is the base, y is the exponent, and x is the argument.

The function f (x) = log b x is read as “log base b of x.” Logarithms are useful in mathematics because they enable us to perform calculations with very large numbers.

How to Solve Logarithmic Functions?

To solve the logarithmic functions, it is important to use exponential functions in the given expression. The natural log or ln is the inverse of e . That means one can undo the other one i.e.

ln (e x ) = x

To solve an equation with logarithm(s), it is important to know their properties.

Properties of logarithmic functions are simply the rules for simplifying logarithms when the inputs are in the form of division, multiplication, or exponents of logarithmic values.

Some of the properties are listed below.

  • Product rule

The product rule of logarithm states the logarithm of the product of two numbers having a common base is equal to the sum of individual logarithms.

⟹ log a  (p q) = log a  p + log a  q.

  • Quotient rule

The quotient rule of logarithms states that the logarithm of the two numbers’ ratio with the same bases is equal to the difference of each logarithm.

⟹ log a  (p/q) = log a  p – log a q

The power rule of logarithm states that the logarithm of a number with a rational exponent is equal to the product of the exponent and its logarithm.

⟹ log a  (p q ) = q log a p

  • Change of Base rule

⟹ log a p = log x p ⋅ log a x

⟹ log q p = log x p / log x q

  • Zero Exponent Rule

Solving Log Function Properties

Other properties of logarithmic functions include:

  • The bases of an exponential function and its equivalent logarithmic function are equal.
  • The logarithms of a positive number to the base of the same number are equal to 1.

log a  a = 1

  • Logarithms of 1 to any base are 0.

log a  1 = 0

  • Log a 0 is undefined
  • Logarithms of negative numbers are undefined.
  • The base of logarithms can never be negative or 1.
  • A logarithmic function with base 10is called a common logarithm. Always assume a base of 10 when solving with logarithmic functions without a small subscript for the base.

Whenever you see logarithms in the equation, you always think of how to undo the logarithm to solve the equation. For that, you use an exponential function . Both of these functions are interchangeable.

The following table tells the way of writing and interchanging the exponential functions and logarithmic functions . The third column tells about how to read both the logarithmic functions.

Let’s use these properties to solve a couple of problems involving logarithmic functions.

Rewrite exponential function 7 2 = 49 to its equivalent logarithmic function.

Given 7 2 = 64.

Here, the base = 7, exponent = 2 and the argument = 49. Therefore, 7 2 = 64 in logarithmic function is;

⟹ log 7 49 = 2

Write the logarithmic equivalent of 5 3 = 125.

exponent = 3;

and argument = 125

5 3 = 125 ⟹ log 5 125 =3

Solve for x in log  3  x = 2

log  3  x = 2 3 2  = x ⟹ x = 9

If 2 log x = 4 log 3, then find the value of ‘x’.

2 log x = 4 log 3

Divide each side by 2.

log x = (4 log 3) / 2

log x = 2 log 3

log x = log 3 2

log x = log 9

Find the logarithm of 1024 to the base 2.

1024 = 2 10

log 2 1024 = 10

Find the value of x in log 2 ( x ) = 4

Rewrite the logarithmic function log 2 ( x ) = 4 to exponential form.

Solve for x in the following logarithmic function log 2 (x – 1) = 5.

Solution Rewrite the logarithm in exponential form as;

log 2 (x – 1) = 5 ⟹ x – 1 = 2 5

Now, solve for x in the algebraic equation. ⟹ x – 1 = 32 x = 33

Find the value of x in log x 900 = 2.

Write the logarithm in exponential form as;

Find the square root of both sides of the equation to get;

x = -30 and 30

But since, the base of logarithms can never be negative or 1, therefore, the correct answer is 30.

Solve for x given, log x = log 2 + log 5

Using the product rule Log b  (m n) = log b  m + log b  n we get;

⟹ log 2 + log 5 = log (2 * 5) = Log   (10).

Therefore, x = 10.

Solve log  x  (4x – 3) = 2

Rewrite the logarithm in exponential form to get;

x 2  = 4x – 3

Now, solve the quadratic equation. x 2  = 4x – 3 x 2  – 4x + 3 = 0 (x -1) (x – 3) = 0

Since the base of a logarithm can never be 1, then the only solution is 3.

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How to Solve Natural Logarithms Problems? (+FREE Worksheet!)

In this blog post, you will learn more about Natural Logarithms and how to solve problems related to natural logarithms.

How to Solve Natural Logarithms Problems? (+FREE Worksheet!)

Related Topics

  • How to Solve Logarithmic Equations
  • How to Evaluate Logarithms
  • Properties of Logarithms

Step by step guide to solve Natural Logarithms

  • A natural logarithm is a logarithm that has a special base of the mathematical constant \(e\), which is an irrational number approximately equal to \(2.71\).
  • The natural logarithm of \(x\) is generally written as ln \(x\), or \(\log_{e}{x}\).

Natural Logarithms – Example 1:

Solve the equation for \(x\): \(e^x=3\)

If \(f(x)=g(x)\),then: \(ln(f(x))=ln(g(x))→ln(e^x)=ln(3) \)

Use log rule: \(\log_{a}{x^b}=b \log_{a}{x}\), then: \(ln(e^x)=x ln(e)→xln(e)=ln(3) \)

\(ln(e)=1\), then: \(x=ln(3) \)

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Natural logarithms – example 2:.

Solve equation for \(x\): \(ln(2x-1)=1\)

Use log rule: \(a=\log_{b}{b^a}\), then: \(1=ln⁡(e^1 )=ln⁡(e)→ln⁡(2x-1)=ln⁡(e)\)

When the logs have the same base: \(\log_{b}{f(x)}=\log_{b}{g(x)}\), then: \(f(x)=g(x)\)

then: \(ln(2x-1)=ln(e)\), then: \(2x-1=e→x=\frac{e+1}{2}\)

Natural Logarithms – Example 3:

Solve the equation for \(x\): \(e^x=5\)

If \(f(x)=g(x)\),then: \(ln(f(x))=ln(g(x))→ln(e^x)=ln(5) \)

Use log rule: \(\log_{a}{x^b}=b \log_{a}{x}\), then: \(ln(e^x)=x ln(e)→xln(e)=ln(5) \)

\(ln(e)=1\), then: \(x=ln(5) \)

Natural Logarithms – Example 4:

Solve equation for \(x\): \(ln(5x-1)=1\)

Use log rule: \(a=\log_{b}{b^a}\), then: \(1=ln⁡(e^1 )=ln⁡(e)→ln⁡(5x-1)=ln⁡(e)\)

then: \(ln(5x-1)=ln(e)\), then: \(5x-1=e→x=\frac{e+1}{5}\)

Exercises to practice Natural Logarithms

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Solve each equation for \(x\)..

  • \(\color{blue}{e^x=3}\)
  • \(\color{blue}{e^x=4}\)
  • \(\color{blue}{e^x=8}\)
  • \(\color{blue}{ln x=6}\)
  • \(\color{blue}{ln (ln x)=5}\)
  • \(\color{blue}{e^x=9}\)
  • \(\color{blue}{ln⁡(2x+5)=4}\)
  • \(\color{blue}{ln(2x-1)=1}\)
  • \(\color{blue}{x=ln 3}\)
  • \(\color{blue}{x=ln 4,x=2ln⁡(2)}\)
  • \(\color{blue}{x=ln 8,x=3ln⁡(2)}\)
  • \(\color{blue}{x=e^6}\)
  • \(\color{blue}{x=e^{e^5}}\)
  • \(\color{blue}{x=ln 9,x=2ln⁡(3)}\)
  • \(\color{blue}{x=\frac{e^4-5}{2}}\)
  • \(\color{blue}{x=\frac{e+1}{2}}\)

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how to solve a problem with log

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Solving logarithmic equations

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  • How to Solve a Logarithmic Equation
  • log ⁡ ( 3 − x ) + log ⁡ ( 4 − 3 x ) − log ⁡ ( x ) = log ⁡ 7 \log(3 - x) + \log (4 - 3x) - \log(x) = \log7 lo g ( 3 − x ) + lo g ( 4 − 3 x ) − lo g ( x ) = lo g 7
  • 2 log ⁡ 3 ( x + 4 ) − log ⁡ 3 ( − x ) = 2 2 \log_3 {(x + 4)} - \log_3 (- x) = 2 2 lo g 3 ​ ( x + 4 ) − lo g 3 ​ ( − x ) = 2
  • log ⁡ 2 x = 2 + 1 2 log ⁡ 2 ( x − 3 ) \log_2x = 2 + {1\over2} \log_2(x - 3) lo g 2 ​ x = 2 + 2 1 ​ lo g 2 ​ ( x − 3 )
  • ( log ⁡ x ) 2 − log ⁡ x 5 = 14 {(\log x)^2 - \log x^5 = 14} ( lo g x ) 2 − lo g x 5 = 14
  • 2 ( log ⁡ 3 n ) 3 − ( log ⁡ 3 n ) 2 = 0 {2 (\log_3n)^3 - (\log_3n)^2 = 0} 2 ( lo g 3 ​ n ) 3 − ( lo g 3 ​ n ) 2 = 0

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Topic Notes

Solving logarithmic equations, rules or laws of logarithms:.

As you know, a logarithm is a mathematical operation that is the inverse of exponentiation. It is expressed by using the abbreviation "log". Before getting into solving logarithmic equations, there are several strategies and "rules" that we must first familiarize ourselves with.

First of all, in order to solve logarithmic equations, just like with polynomials, you should be comfortable graphing logarithmic functions. Check out our video on graphing logarithmic functions for an overview if needed. Also, before we get into logarithm rules, it is important that you also understand one of the simplest logarithm strategies – the change of base formula. Again, check out our video on the change of base formula if you need a refresher. Now that you have all that mastered, let's take a look at some of the most important logarithm rules:

1) Logarithm Product Rule

In general, the product rule of logarithms is defined by:

log ⁡ A + log ⁡ B = log ⁡ ( A × B ) \log A + \log B = \log (A \times B) lo g A + lo g B = lo g ( A × B )

That is, when adding two logs of the same base , you can rewrite the expression as a single log by multiplying the terms within the logarithmic expression.

2) Logarithm Quotient Rule

In general, the quotient rule of logarithms is defined by:

log ⁡ A − log ⁡ B = log ⁡ ( A B ) \log A - \log B = \log (\frac{A}{B}) lo g A − lo g B = lo g ( B A ​ )

That is, when subtracting two logs of the same base , you can rewrite the expression as a single log by dividing the terms within the logarithmic expression.

3) Logarithm Power Rule

In general, the power rule of logarithms is defined by:

log ⁡ ( A ) B = B × log ⁡ A \log (A)^{B} = B \times \log A lo g ( A ) B = B × lo g A

That is, when there is an exponent on the term within the logarithmic expression, you can bring down that exponent and multiply it by the log.

4) Log of Exponent Rule

In general, the log of exponent rule is defined by:

log ⁡ A ( A B ) = B \log_{A} (A^{B}) = B lo g A ​ ( A B ) = B

That is, when there is an exponent on the term within the logarithmic expression, and that term is the same as the base of the logarithm, the answer is simply the exponent.

5) Exponent of Log Rule

In general, the exponent of log rule is defined by:

A log ⁡ A ( B ) = B A^{\log_{A} (B)} = B A l o g A ​ ( B ) = B

That is, raising a logarithm of a number by its base equals that number.

6) Log Identity Rule

In general, the identity rule of logarithms is defined by:

log ⁡ A A = 1 \log_{A} A = 1 lo g A ​ A = 1

That is, when taking the log of something to the base of that same thing, the logarithmic expression is simply equal to just 1.

7) Special Logs

Though not necessarily rules, there are a couple of logs that you should know by heart to make things a little easier. They are:

log ⁡ 1 = 0 \log 1 = 0 lo g 1 = 0

log ⁡ 0 = u n d e f i n e d \log 0 = undefined lo g 0 = u n d e f in e d

Both of these cases are always true, regardless of the base. Also, in case it comes up, the first special case is sometimes referred to as the logarithmic zero rule.

All of these rules, taken together, are extremely powerful tools we can use to solve any logarithmic problem. For a video review of these concepts, check out our videos on properties of logarithms and the quotient rule for logarithms . Now that we've covered the essentials, let's get to how to solve log problems!

How to Solve Log Problems:

As with anything in mathematics, the best way to learn how to solve log problems is to do some practice problems! We will use the rules we have just discussed to solve some examples.

Solve the logarithmic equation:

log ⁡ ( 3 − x ) + log ⁡ ( 4 − 3 x ) − log ⁡ ( x ) = log ⁡ 7 \log (3 - x) + \log (4 - 3x) - \log (x) = \log 7 lo g ( 3 − x ) + lo g ( 4 − 3 x ) − lo g ( x ) = lo g 7

Step 1: Use Known Log Rules

In any problem that involves solving logarithmic equations, the first step is to always try to simplify using the log rules. In this case, we will use the product, quotient, and exponent of log rules. We do this to try to make a polynomial/algebraic equation that is easier to solve. This is shown below:

log ⁡ [ ( 3 − x ) ( 4 − 3 x ) x ] = log ⁡ 7 \log [\frac{(3-x)(4-3x)}{x}] = \log 7 lo g [ x ( 3 − x ) ( 4 − 3 x ) ​ ] = lo g 7

1 0 log ⁡ [ ( 3 − x ) ( 4 − 3 x ) x ] = 1 0 log ⁡ 7 10^{\log [\frac{(3-x)(4-3x)}{x}]} = 10^{\log 7} 1 0 l o g [ x ( 3 − x ) ( 4 − 3 x ) ​ ] = 1 0 l o g 7

( 3 − x ) ( 4 − 3 x ) x = 7 \frac{(3-x)(4-3x)}{x} = 7 x ( 3 − x ) ( 4 − 3 x ) ​ = 7

Step 2: Solve Equation

We are left with an algebraic equation which we can now solve.

3 x 2 − 13 x + 12 x = 7 \frac{3x^{2} - 13x + 12}{x} = 7 x 3 x 2 − 13 x + 12 ​ = 7

3 x 2 − 13 x + 12 = 7 x 3x^{2} - 13x + 12 = 7x 3 x 2 − 13 x + 12 = 7 x

3 x 2 − 20 x + 12 = 0 3x^{2} - 20x + 12 = 0 3 x 2 − 20 x + 12 = 0

( 3 x − 2 ) ( x − 6 ) = 0 (3x - 2)(x - 6) = 0 ( 3 x − 2 ) ( x − 6 ) = 0

3 x − 2 = 0 o r x − 6 = 0 3x - 2 = 0 or x - 6 = 0 3 x − 2 = 0 or x − 6 = 0

x = 2 3 o r x = 6 x = \frac{2}{3} or x = 6 x = 3 2 ​ or x = 6

Step 3: Check Solutions

Because we initially had a logarithmic equation, we need to check our answers to make sure they are valid.

log ⁡ ( 3 − 2 3 ) + log ⁡ ( 4 − 3 ( 2 3 ) ) − log ⁡ ( 2 3 ) = log ⁡ 7 \log (3 - \frac{2}{3}) + \log (4 - 3(\frac{2}{3})) - \log (\frac{2}{3}) = \log 7 lo g ( 3 − 3 2 ​ ) + lo g ( 4 − 3 ( 3 2 ​ )) − lo g ( 3 2 ​ ) = lo g 7

The solution x = 2 3 x = \frac{2}{3} x = 3 2 ​ is correct.

log ⁡ ( 3 − 6 ) + log ⁡ ( 4 − 3 ( 6 ) ) − log ⁡ ( 6 ) = log ⁡ 7 \log (3 - 6) + \log (4 - 3(6)) - \log (6) = \log 7 lo g ( 3 − 6 ) + lo g ( 4 − 3 ( 6 )) − lo g ( 6 ) = lo g 7

The solution x = 6 x = 6 x = 6 is rejected because the log of a negative number is undefined.

In this case, we will use the power and quotient log rules. We do this to try to make a polynomial/algebraic equation that is easier to solve. This is shown below:

2 log ⁡ 3 ( x + 4 ) − log ⁡ 3 ( − x ) = 2 2\log_{3} (x + 4) - \log_{3} (-x) = 2 2 lo g 3 ​ ( x + 4 ) − lo g 3 ​ ( − x ) = 2

log ⁡ 3 ( x + 4 ) 2 − log ⁡ 3 ( − x ) = 2 \log_{3} (x + 4)^{2} - \log_{3} (-x) = 2 lo g 3 ​ ( x + 4 ) 2 − lo g 3 ​ ( − x ) = 2

log ⁡ 3 ( x 2 + 8 x + 16 − x ) = 2 \log_{3} (\frac{x^{2} + 8x + 16}{-x}) = 2 lo g 3 ​ ( − x x 2 + 8 x + 16 ​ ) = 2

Step 2: Simplify

We can convert to exponent form because one side has log and the other side does not.

3 2 = x 2 + 8 x + 16 − x 3^{2} = \frac{x^{2} + 8x + 16}{-x} 3 2 = − x x 2 + 8 x + 16 ​

Step 3: Solve Equation

− 9 x = x 2 + 8 x + 16 -9x = x^{2} + 8x + 16 − 9 x = x 2 + 8 x + 16

x 2 + 17 x + 16 = 0 x^{2} + 17x + 16 = 0 x 2 + 17 x + 16 = 0

( x + 16 ) ( x + 1 ) = 0 (x + 16)(x + 1) = 0 ( x + 16 ) ( x + 1 ) = 0

x = − 16 o r x = − 1 x = -16 or x = -1 x = − 16 or x = − 1

Step 4: Check Solutions

2 log ⁡ 3 ( ( − 16 ) + 4 ) − log ⁡ 3 ( − ( − 16 ) ) = 2 2\log_{3} ((-16) + 4) - \log_{3} (-(-16)) = 2 2 lo g 3 ​ (( − 16 ) + 4 ) − lo g 3 ​ ( − ( − 16 )) = 2

The solution x = -16 is rejected.

2 log ⁡ 3 ( ( − 1 ) + 4 ) − log ⁡ 3 ( − ( − 1 ) ) = 2 2\log_{3} ((-1) + 4) - \log_{3} (-(-1)) = 2 2 lo g 3 ​ (( − 1 ) + 4 ) − lo g 3 ​ ( − ( − 1 )) = 2

The solution x = -1 is correct.

Step 1: Simplify

Multiply both sides of the equation by 2 to get rid of the fraction.

log ⁡ 2 x = 2 + 1 2 log ⁡ 2 ( x − 3 ) \log_{2} x = 2 + \frac{1}{2} \log_{2} (x - 3) lo g 2 ​ x = 2 + 2 1 ​ lo g 2 ​ ( x − 3 )

2 log ⁡ 2 x = 4 + log ⁡ 2 ( x − 3 ) 2\log_{2} x = 4 + \log_{2} (x - 3) 2 lo g 2 ​ x = 4 + lo g 2 ​ ( x − 3 )

Step 2: Use Known Log Rules

In this case, we will use the power of log and quotient log rules. We can then simplify like in the previous example to make the exponential form. We do this to try to make a polynomial/algebraic equation that is easier to solve. This is shown below:

log ⁡ 2 ( x ) 2 = 4 + log ⁡ 2 ( x − 3 ) \log_{2} (x)^{2} = 4 + \log_{2} (x - 3) lo g 2 ​ ( x ) 2 = 4 + lo g 2 ​ ( x − 3 )

log ⁡ 2 ( x ) 2 − log ⁡ 2 ( x − 3 ) = 4 \log_{2} (x)^{2} - \log_{2} (x - 3)= 4 lo g 2 ​ ( x ) 2 − lo g 2 ​ ( x − 3 ) = 4

log ⁡ 2 ( x 2 x − 3 ) = 4 \log_{2} (\frac{x^{2}}{x-3}) = 4 lo g 2 ​ ( x − 3 x 2 ​ ) = 4

2 4 = x 2 x − 3 2^{4} = \frac{x^{2}}{x-3} 2 4 = x − 3 x 2 ​

16 = x 2 x − 3 16 = \frac{x^{2}}{x-3} 16 = x − 3 x 2 ​

16 x − 48 = x 2 16x - 48 = x^{2} 16 x − 48 = x 2

x 2 − 16 x + 48 = 0 x^{2} - 16x + 48 = 0 x 2 − 16 x + 48 = 0

( x − 4 ) ( x − 12 ) = 0 (x - 4)(x - 12) = 0 ( x − 4 ) ( x − 12 ) = 0

x = 4 o r x = 12 x = 4 or x = 12 x = 4 or x = 12

log ⁡ 2 4 = 2 + 1 2 log ⁡ 2 ( 4 − 3 ) \log_{2} 4 = 2 + \frac{1}{2} \log_{2} (4 - 3) lo g 2 ​ 4 = 2 + 2 1 ​ lo g 2 ​ ( 4 − 3 )

The solution x = 4 checks out.

log ⁡ 2 12 = 2 + 1 2 log ⁡ 2 ( 12 − 3 ) \log_{2} 12 = 2 + \frac{1}{2} \log_{2} (12 - 3) lo g 2 ​ 12 = 2 + 2 1 ​ lo g 2 ​ ( 12 − 3 )

So does x=12. In this problem, we get to keep both our answers.

In this case, we will use the exponent of log rule. We do this to try to make a polynomial/algebraic equation that is easier to solve. Note: ( log ⁡ x ) 2 (\log x)^{2} ( lo g x ) 2 is different than log ⁡ x 2 \log x^{2} lo g x 2 , and thus we cannot simplify the first log ⁡ \log lo g .This is shown below:

( log ⁡ x ) 2 − log ⁡ x 5 = 14 (\log x)^{2} - \log x^{5} = 14 ( lo g x ) 2 − lo g x 5 = 14

( log ⁡ x ) 2 − 5 log ⁡ x = 14 (\log x)^{2} - 5\log x = 14 ( lo g x ) 2 − 5 lo g x = 14

Step 2: Substitution

To make this equation easier to solve, we can substitute log ⁡ x \log x lo g x as "a" to make a quadratic equation!

a 2 − 5 a = 14 a^{2} - 5a = 14 a 2 − 5 a = 14

a 2 − 5 a − 14 = 0 a^{2} - 5a - 14 = 0 a 2 − 5 a − 14 = 0

( a − 7 ) ( a + 2 ) = 0 (a - 7)(a + 2) = 0 ( a − 7 ) ( a + 2 ) = 0

a = 7 o r a = − 2 a = 7 or a = -2 a = 7 or a = − 2

*** Since we used substitution, we need to replace "a" back with the original term! ***

log ⁡ x = 7 o r log ⁡ x = − 2 \log x = 7 or \log x = -2 lo g x = 7 or lo g x = − 2

1 0 7 = x o r 1 0 − 2 = x 10^{7} = x or 10^{-2} = x 1 0 7 = x or 1 0 − 2 = x

( log ⁡ 1 0 7 ) 2 − log ⁡ ( 1 0 7 ) 5 = 14 (\log 10^{7})^{2} - \log (10^{7})^{5} = 14 ( lo g 1 0 7 ) 2 − lo g ( 1 0 7 ) 5 = 14

The solution x = 1 0 7 x = 10^{7} x = 1 0 7 is correct.

( log ⁡ 1 0 − 2 ) 2 − log ⁡ ( 1 0 − 2 ) 5 = 14 (\log 10^{-2})^{2} - \log (10^{-2})^{5} = 14 ( lo g 1 0 − 2 ) 2 − lo g ( 1 0 − 2 ) 5 = 14

The solution x = 1 0 − 2 x = 10^{-2} x = 1 0 − 2 is not correct.

And that's all there is too it! To check your work with future practice problems, be sure to use this excellent calculator here . Lastly, for a video review of everything we've just covered, check out our video on how to solve log equations .

Basic Concepts

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Google pixel fold: common problems and how to solve them.

Here's how to resolve some of the biggest problems with Google's first foldable

The Google Pixel Fold does many things well, but it also has problems, so it struggles to compete with other popular foldable Android phones. However, you can solve many of these problems and turn the problematic phone into a great one.

We walk you through the biggest problems with the Pixel Fold and how to solve or mitigate them. We recommend exploring all the best features of the Pixel Fold , as its unique elements help it stand out in the ever-narrowing crowd of Android phones.

Apps don't scale properly in horizontal mode

One of the biggest and most noticeable problems with the Pixel Fold is the inconsistent scaling applied to apps. While most of Google's apps are designed to fit the Pixel Fold's outer and inner screens in horizontal or landscape mode, some popular apps, including Instagram and Yelp, don't.

The only "fix" for this is to keep your Pixel Fold in portrait mode. Portrait mode doesn't create as many errors, thanks to the phone's unique aspect ratio. We recommend keeping landscape mode for apps like Calendar or Drive or for watching media.

instragram on pixel fold in horizontal mode

Battery charges slow and doesn't last long

Many users have noted the Pixel Fold's poor charging speeds and battery life. The Pixel Fold has a max charging speed of 21W, so you can't solve this by buying a new charging brick. However, there are a few ways you can improve its battery life.

You can increase the phone's battery life by activating adaptive battery, turning off smooth display, and adjusting your Pixel Fold's screen brightness. We have a detailed guide on all the settings you can tweak to improve the Pixel Fold's battery life.

The inner screen turns on when the Pixel Fold is closed

When closed, pressing the Pixel Fold's power button might turn on the inner screen alongside the outer screen. This is not an issue with the Pixel Fold, but with some cases that use magnets in their design. These magnets interfere with the inner screen, causing it to turn on unexpectedly alongside other problems.

The only way to resolve this is to buy a case without magnets. There are plenty of great Pixel Fold cases that don't use magnets. Double-check before you buy, as they often don't clearly state whether they use magnets.

Permanent damage to the inner screen

Complaints about the durability of the Pixel Fold were widespread even before it launched, with some reviewers noting the phone's inner screen broke after a few days of use. If your Pixel Fold arrived broken or has other hardware issues, contact Google Support for a replacement or fix. We recommend taking preventative steps to avoid it breaking in the first place.

Many of the inner screen issues stem from a small gap in the hinge, which traps particles between the Pixel Fold's bezel and screen protector. This is an issue across foldables. Buy a case that doesn't leave a gap around the hinge (like the spine of a hardback book) that can minimize the amount of dust entering the phone.

Hinge doesn't lie flat

The Pixel Fold stops a few degrees short of lying flat when opened. If you apply force, you can make it lie completely flat for a while, but we don't recommend doing this. After some time with the Fold, we found that the phone became more resistant to lying flat the longer we used it. Applying more force can cause issues with the hinge.

Don't apply additional pressure when opening the Pixel Fold. One of the best things about the phone is that the hinge holds it tight at any angle apart from flat. Open it to its limit and leave it there.

Unresponsive outer screen

The inner and outer screens of the Pixel Fold aren't identical. While the outer screen is higher quality than the inner, users have noted that it can become unresponsive. Opening and closing the phone is a temporary fix. You can also tweak some settings for a permanent solution.

  • Open your Pixel Fold's Settings app.
  • Scroll down and tap Display .

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If this doesn't solve the problem, follow Google's official guide to fix a Pixel screen that isn't working correctly.

Apps disappear when opening the Pixel Fold

If two apps are open simultaneously on the Pixel Fold's inner screen, closing the phone causes one app to disappear when you reopen it. This is intended behavior. The Pixel Fold only keeps the active app open on the inner screen when the phone is closed.

There is no way to resolve this problem. You can minimize the frustration by tapping the more important app once before closing the phone. This ensures it stays open when reopening the phone.

Internal temperature becomes uncomfortably hot

The Pixel Fold is more prone to overheating than any other Pixel phone. This is a design flaw. There isn't a way to permanently stop it from growing uncomfortably warm. However, you can do a few things to keep the internal temperature low. If you're charging your Pixel Fold or are in a warm environment, don't perform these actions:

  • Take high-resolution photographs or videos.
  • Watch a video over a mobile connection.
  • Multitask with intensive apps.

Don't let the Pixel Fold ruin your view of foldables

The Pixel Fold suffers from many problems that aren't common to all foldables. If your experience with the Pixel Fold is too frustrating, we recommend exploring the other great Android foldables .

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Baldur's gate 3: vlaakith guide (should you obey).

In Baldur's Gate 3, players can potentially find a deadly adversary in the githyanki Queen Vlaakith. Knowing when to follow her commands is crucial.

Vlaakith, the undying Lich Queen of the Githyanki, is a potentially deadly adversary in Baldur’s Gate 3 , depending on whether the party obeys or disobeys her commands. The leader of the Githyanki is just one of the many gods and goddesses players can encounter. However, the decisions made when interacting with Vlaakith can shape the fate of your companion, Lae’zel, and the game's ending.

[Warning: The following article contains spoilers for Baldur's Gate 3.] When journeying through Act One of Baldur’s Gate 3 , the party will have the opportunity to visit the Githyanki Crèche Y’llek in Rosymorn Monastery. The Githyanki may have a cure for mind flayer parasites, and players can opt to negotiate with them to find out more. However, it’s also possible to skip Creche Y’llek entirely and miss out on an encounter with Vlaakith herself.

Lae'zel and other githyanki in the background in BG3.

Baldur's Gate 3: You Shouldn't Skip The Githyanki Crèche Questline

Should you obey vlaakith in act one of baldur’s gate 3.

After failing to find a cure at Crèche Y’llek, players can seek out Inquisitor Ch’r’ai W’wargaz , who is visiting the Crèche to further his search for the Mysterious Artefact. The artefact in question is the Astral Prism in the party’s possession, which protects them from being dominated by the Absolute. When meeting the Inquisitor, you can try to hand over the Prism or battle to keep it for yourself.

The battle against the Githyanki Inquisitor can be challenging, so make sure to take a Short or Long Rest before continuing.

Regardless of what you do, Vlaakith will appear as a massive projection and demand that the party enter the Prism to slay its inhabitant: the Dream Visitor and Guardian. You can greet her and show deference or wave (although doing so earns disapproval from Lae’zel). In summary, you should obey Vlaakith’s command to enter the Prism. At this stage, disobeying Vlaakith by refusing or asking her to do it herself or showing her too much disrespect will cause her to cast a spell that kills the whole party instantly. This leads to a Game Over screen.

This is one of the worst decisions you can make in BG3 's Honor Mode , as it’s guaranteed to end your Honor Mode run.

However, once you've met the Dream Visitor inside the Prism, you can leave them alone or attempt to kill them . It's worth noting that trying to kill the Dream Visitor has no effect, as they simply disappear. At this point, you can exit the Prism to find that Vlaakith has betrayed Lae'zel and branded her a traitor. Engaging with Vlaakith has consequences for Lae'zel's companion quest, The Githyanki Warrior , which leads into Acts Two and Three.

If players don’t visit Crèche Y’llek, Lae’zel will continue to be loyal to Vlaakith, who only appears in Act Three.

Should You Side With Kith’rak Voss In Baldur’s Gate 3?

Kith'rak Voss Talking To Player In Sewers In Baldur's Gate 3

If Vlaakith betrays Lae’zel, your party camp will be visited by Kith’rak Voss during your next Long Rest. Voss will attempt to convince Lae’zel that Vlaakith is a usurper who stole the throne from Orpheus, the Githyanki prince and rightful heir. Voss will head to Baldur’s Gate following your conversation.

The party can encounter him later in Act Three, attempting to bargain with the devil Raphael in Sharess' Caress. This conversation is important, as it can cause Lae'zel to start doubting her faith in Vlaakith.

Should You Obey Vlaakith Or Follow Voss And Orpheus In Act Three of Baldur’s Gate 3?

Lae'zel Gazes Up At Blue Projection Of Vlaakith In Baldur's Gate 3.

Once players start Act Three of Baldur’s Gate 3 , Vlaakith will appear over your camp once you Long Rest in Rivington. At this point, she demands that Lae’zel slay Orpheus, promising to make Lae’zel the commander of her armies if she does so. Your decisions here can influence Lae’zel to declare her support for Orpheus or stay loyal to Vlaakith, and it is a significant turning point in her companion quest.

You can obey Vlaakith or choose to side with Orpheus, and your decision will change Lae’zel’s fate at the end of the game. Generally speaking, you shouldn't force Lae'zel to obey Vlaakith. If she goes against Vlaakith, Lae'zel leaves for the Astral Plane at the end of the game and leads a rebellion to save her people. If she sides with Vlaakith, Withers implies that she meets a much darker fate during the epilogue party.

Persuading Lae’zel to side with Vlaakith requires passing a tough Persuasion check , which seems to imply that Lae’zel’s canonical path is to side with Orpheus and Voss.

If she sides with Voss and Orpheus, Lae'zel will want to enter the House of Hope to find the Orphic Hammer to free Orpheus; if not, she will want the Hammer so she can free him and then kill him. Ultimately, players will have to choose whether to side with Orpheus or the Emperor :

  • The party can free Orpheus and side with him instead of the Emperor. However, the Emperor will join the Absolute, and either the player character, Orpheus, or Karlach will have to become a mind flayer to defeat the Absolute.
  • The party can refuse to free Orpheus, allowing the Emperor to consume him to gain his abilities.

Your choice will have lasting consequences on Baldur's Gate 3's ending. However, as long as your characters obey Vlaakith during your first encounter, you can deal with her however you like in subsequent interactions.

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Ap decision notes: what to expect in bridgeport’s do-over mayoral election.

WASHINGTON (AP) — If it's Tuesday, there must be a mayoral election in Bridgeport. For the fourth time in five ...

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A watch that melted during the atomic blast over Hiroshima, Japan, sells for more than $31,000

AI helps boost creativity in the workplace but still can't compete with people's problem-solving skills, study finds

  • Artificial intelligence is disrupting professional workplaces with systems like ChatGPT and Gemini.
  • A study found that people mistrust AI for the wrong reasons while trusting it for tasks where it might mess up.
  • AI boosts performance in creative tasks but performs poorly in problem-solving, the study found.

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Artificial intelligence is coming to change your workplace .

The rapidly evolving technology has already started to disrupt day-to-day activities in professional settings, and leaders at the forefront of the AI revolution have been clear about how they hope to implement systems like ChatGPT and Gemini into the mainstream workflow.

But while many employees may be cautiously skeptical about the impending AI overhaul, a recent study found that people are actually mistrustful of artificial intelligence for the wrong reasons while frequently trusting in the technology for tasks it's more likely to mess up.

The September 2023 study, which is titled " How People Can Create—and Destroy—Value with Generative AI ," was spearheaded by François Candelon, the managing director and senior partner at consulting company Boston Consulting Group.

The study's findings are back in the news this week after Candelon sat down with the Wall Street Journal's Executive Insights podcast to discuss generative AI in the workplace.

Candelon partnered with talent from top universities like MIT, Wharton, Harvard Business School, and the University of Warwick, and used his consulting company's own employees to execute the experiment, which he told The Journal was inspired by his desire to figure out how humans and AI can work together to help businesses.

The more than 750 study participants were given real tasks, including "creative product innovation" assignments. The participants were instructed to use OpenAI tool GPT-4 to help them with tasks like pitching the shoe concepts to their boss, coming up with focus group questions, and executing a successful social media rollout, Candelon said.

The study found that people using AI faired much better than those working without it when it came to creative product innovation tasks. About 90% of the participants improved their performance when using AI for any task involving ideation and content creation.

Participants also converged on a performance level that was 40% higher than those working on the same task without GPT-4, according to the study.

The most benefits were seen when people didn't try to change or improve the technology's output suggestions, accepting GPT-4's suggestions as is, the study found.

But there are still some tasks where humans have the edge. People's problem-solving skills far outweigh the help offered by AI, Candelon said.

The study found that generative AI actually persuaded several participants to accept GPT's misleading output, even when they had been briefed on the possibility of wrong answers.

Participants who used AI for problem-solving tasks performed 23% worse than those who didn't use the tool at all, according to the study.

The "double-edged sword" that is generative AI, with its "relatively uniform output," can also reduce a group's diversity of thought by 41%, the study found.

But Candelon stressed to The Journal that AI is exceedingly powerful and, ultimately, unavoidable.

"There is this famous quote saying that humans won't get replaced by AI. They will get replaced by humans using AI," he told the outlet.

Candelon said the study shows that data will become even more important with generative AI in the workplace, forcing people to revisit their workflows and figure out places for human and AI collaboration.

how to solve a problem with log

Watch: AI will drive personalization, not creativity, says Roku's VP of growth marketing, Sweta Patel

how to solve a problem with log

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how to solve a problem with log

ANALYSIS: Can AI Solve ESG’s Data Problem?

By Abigail Gampher Takacs

Abigail Gampher Takacs

Artificial intelligence disclosures to the SEC are on the rise . The references to AI are many and varied; some companies use it to cut costs, while others discuss AI’s risks.

A closer look at SEC filings in the context of ESG shows how AI could affect ESG data collection and disclosure. This is important because stakeholders are increasingly requesting that companies publish ESG data—including everything from information on their workforce demographics to their carbon footprint.

However, the disclosure of accurate and comparable ESG data has hit a major roadblock : Getting ESG data ready for public disclosure requires companies ...

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COMMENTS

  1. Solving Logarithmic Equations

    Example 1: Solve the logarithmic equation. Since we want to transform the left side into a single logarithmic equation, we should use the Product Rule in reverse to condense it. Here is the rule, just in case you forgot. Given Apply Product Rule from Log Rules. Distribute: [latex]\left ( {x + 2} \right)\left ( 3 \right) = 3x + 6 [/latex]

  2. 3 Ways to Solve Logarithms

    1 Know the logarithm definition. Before you can solve logarithms, you need to understand that a logarithm is essentially another way to write an exponential equation. It's precise definition is as follows: y = logb (x) If and only if: by = x Note that b is the base of the logarithm. It must also be true that: b > 0 b does not equal 1

  3. Intro to Logarithms (article)

    log b ( a) = c b c = a Both equations describe the same relationship between a , b , and c : b is the base , c is the exponent , and a is called the argument . A helpful note

  4. Logarithms

    Familiar Attempted Not started Quiz Unit test About this unit Logarithms are the inverses of exponents. They allow us to solve challenging exponential equations, and they are a good excuse to dive deeper into the relationship between a function and its inverse. Introduction to logarithms Learn Intro to logarithms Intro to Logarithms

  5. Logarithmic Equation Calculator

    To solve a logarithmic equations use the esxponents rules to isolate logarithmic expressions with the same base. Set the arguments equal to each other, solve the equation and check your answer. What is logarithm equation? A logarithmic equation is an equation that involves the logarithm of an expression containing a varaible.

  6. Algebra

    Show All Solutions Hide All Solutions a 2log9(√x)−log9(6x −1) = 0 2 log 9 ( x) − log 9 ( 6 x − 1) = 0 Show Solution b logx +log(x−1) =log(3x+12) log x + log ( x − 1) = log ( 3 x + 12) Show Solution c ln10−ln(7 −x) = lnx ln 10 − ln ( 7 − x) = ln x Show Solution

  7. Evaluating logarithms (advanced) (video)

    . It seems in these problems that you just have to recognize roots to solve these logs without a calculator or one is doomed to trial and error. How does one solve LOG problems, without calculator or prefigured tables, on numbers that really don't come out nicely like LOG base 3 of 13? • ( 107 votes) Upvote ArDeeJ 11 years ago One doesn't, really.

  8. Algebra

    Solve each of the following equations. log4(x2−2x) = log4(5x −12) log 4 ( x 2 − 2 x) = log 4 ( 5 x − 12) Solution log(6x) −log(4−x) =log(3) log ( 6 x) − log ( 4 − x) = log ( 3) Solution ln(x)+ln(x+3) =ln(20−5x) ln ( x) + ln ( x + 3) = ln ( 20 − 5 x) Solution log3(25−x2) = 2 log 3 ( 25 − x 2) = 2 Solution

  9. How To Solve Logarithmic Equations

    What is the general strategy for solving log equations? Example 1 Logarithm on one side and a number on the other General method for solving this type (log on one side), Rewrite the logarithm as exponential equation and solve. Let's look at a specific example: log4x + log48 = 3 l o g 4 x + l o g 4 8 = 3 Step 1 Rewrite log side as single logarithm

  10. Using laws of logarithms (laws of logs) to solve log problems

    Laws of logarithms (or laws of logs) include product, quotient, and power rules for logarithms, as well as the general rule for logs (and the change of base formula we'll cover in the next lesson), can all be used together, in any combination, in order to solve log problems.

  11. 4.7: Exponential and Logarithmic Equations

    Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form bS = bT. Use the one-to-one property to set the exponents equal. Solve the resulting equation, S = T, for the unknown. Example 4.7.1: Solving an Exponential Equation with a Common Base. Solve 2x − 1 = 22x − 4.

  12. Solving Log Equations from the Definition

    Solving Log Equations from the Definition Using Exponentials Calculators & Etc. Purplemath The first type of logarithmic equation has two logs, each having the same base, which have been set equal to each other.

  13. Introduction to Logarithms

    Sometimes a logarithm is written without a base, like this: log (100) This usually means that the base is really 10. It is called a "common logarithm". Engineers love to use it. On a calculator it is the "log" button. It is how many times we need to use 10 in a multiplication, to get our desired number. Example: log (1000) = log10(1000) = 3.

  14. Log problems: pH, decibels, and the Richter Scale

    The formula for the loudness of a sound is given by dB = 10×log [I ÷ I0] where I0 is the intensity of "threshold sound", or sound that can barely be perceived. Other sounds are defined in terms of how many times more intense they are than threshold sound.

  15. How to Solve a Logarithmic Equation (Guide)

    Step 1: Use the properties of logarithms - Begin by isolating the logarithmic expression on one side of the equation. Apply the rules of logarithms, such as the product rule, quotient rule, and power rule, to simplify the equation.

  16. Solving Logarithmic Functions

    12 2 = 144. log 12 144 = 2. log base 12 of 144. Let's use these properties to solve a couple of problems involving logarithmic functions. Example 1. Rewrite exponential function 7 2 = 49 to its equivalent logarithmic function. Solution. Given 7 2 = 64. Here, the base = 7, exponent = 2 and the argument = 49.

  17. How to Solve Natural Logarithms Problems? (+FREE Worksheet!)

    Natural Logarithms - Example 2: Solve equation for \(x\): \(ln(2x-1)=1\) Solution: Use log rule: \(a=\log_{b}{b^a}\), then: \(1=ln⁡(e^1 )=ln⁡(e)→ln⁡(2x-1 ...

  18. How to solve logarithmic equations

    Step 1: Use Known Log Rules. In any problem that involves solving logarithmic equations, the first step is to always try to simplify using the log rules. In this case, we will use the product, quotient, and exponent of log rules. We do this to try to make a polynomial/algebraic equation that is easier to solve. This is shown below: log ⁡ [(3 ...

  19. Solve

    QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students. The algebra section allows you to expand, factor or simplify virtually any expression you choose. It also has commands for splitting fractions into partial fractions, combining several fractions into one and ...

  20. Microsoft Math Solver

    Get math help in your language. Works in Spanish, Hindi, German, and more. Online math solver with free step by step solutions to algebra, calculus, and other math problems. Get help on the web or with our math app.

  21. Logarithmic equations: variable in the argument

    Marcos Vinicius 11 years ago The human ear works as a logarithmic function. The tempered musical scale is exponential so after passing through a logarithmic function (ear) it become linear. This mix of functions makes the transition from notes of the scale perceived by our brain softly as if the notes were located exactly one after the other.

  22. Step-by-Step Calculator

    Full pad Examples Frequently Asked Questions (FAQ) Is there a step by step calculator for math? Symbolab is the best step by step calculator for a wide range of math problems, from basic arithmetic to advanced calculus and linear algebra. It shows you the solution, graph, detailed steps and explanations for each problem.

  23. Save 30% on CRASH MATH on Steam

    You must solve math problems while trying to drive a car and avoid traps. This driving game has a lot of possibilities. The physics engine simulates every component of a vehicle in real-time, resulting in a funny behavior. While creating the excitement of solving math problems.

  24. Google Pixel Fold: Common problems and how to solve them

    Battery charges slow and doesn't last long . Many users have noted the Pixel Fold's poor charging speeds and battery life. The Pixel Fold has a max charging speed of 21W, so you can't solve this ...

  25. Nuclear fusion: Scientists say they can use AI to solve a key problem

    On Wednesday, researchers from Princeton University and the Princeton Plasma Physics Laboratory reported in the journal Nature they found a way to use AI to forecast these potential instabilities ...

  26. Baldur's Gate 3: Vlaakith Guide (Should You Obey?)

    How To Solve The Chamber of Justice Puzzle In Baldur's Gate 3 The Wyrmway in the Lower City of Baldur's Gate 3 hosts a complex puzzle in the Chamber of Justice that you must solve through your sense of judgment.

  27. Vermont governor signs school funding bill but says it won't solve

    SIGN UP TO CONTINUEPrint Subscriber? Sign Up for Full Access!Please sign up for as low as 36 cents per day to continue viewing our website.Digital subscribers receiveUnlimited access to all stories from nashuatelegraph.com on your computer, tablet or smart phone.Access nashuatelegraph.com, view our digital edition or use our Full Access apps.Get more information at nashuatelegraph.com ...

  28. AI Boosts Creativity, Can't Match People's Problem-Solving Skills: Study

    A recent study on generative AI in the workplace found that the technology can actually lead people to make mistakes when it comes to problem-solving. Menu icon A vertical stack of three evenly ...

  29. ANALYSIS: Can AI Solve ESG's Data Problem?

    Artificial intelligence disclosures to the SEC are on the rise. The references to AI are many and varied; some companies use it to cut costs, while others discuss AI's risks. A closer look at SEC filings in the context of ESG shows how AI could affect ESG data collection and disclosure. This is ...