solve algebra exponent problems

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Mathematics
Exponents and Logarithms

How to Solve Algebraic Problems With Exponents

Last Updated: May 2, 2023 References

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. This article has been viewed 71,018 times.

In algebra, the operations (adding, subtracting, multiplying, and dividing) performed on variables work the same as the operations performed on numbers. When performing these operations on exponents, however, the laws are different. By learning these special rules for exponents, you can easily simplify algebraic expressions that include them.

Solving a Problem with Exponents

Image titled Solve Algebraic Problems With Exponents Step 6

Understanding the Laws of Exponents

Image titled Solve Algebraic Problems With Exponents Step 1

Video . By using this service, some information may be shared with YouTube.

$-x{^{2}}=-1(x\times x)=-x$

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How to Solve an Exponential Equation

How to solve an exponential equation: video lesson, solving an exponential equation: example 1.

Solve 5 x = 13.

Step 1. Take logs of both sides

5 x = 13 becomes log(5 x ) = log(13).

Step 2. Bring down the power in front of the log

The power of x can be written in front of the log so that log(5 x ) = log(13) becomes xlog(5) = log(13).

Step 3. Solve the resulting equation for x

xlog(5) = log(13) can be solved for x by dividing both sides of the equation by log(5).

This can be evaluated on a calculator to get 𝑥 ≈ 1.59.

Solving an Exponential Equation: Example 2

Solve 3 2x = 0.51

3 2x = 0.51 becomes log(3 2x ) = log(0.51).

The power of 2x can be written in front of the log so that log(3 2x ) = log(0.51) becomes 2xlog(3) = log(0.51).

how to solve an exponential equation with a negative answer

2xlog(3) = log(0.51) can be solved by dividing both sides by 2log(3).

This can be evaluated as 𝑥 ≈ -0.306.

How to Solve Exponential Equations with Different Bases

For example solve the exponential equation 5 𝑥 = 2 𝑥+2 .

Step 1. Take logarithms of both sides

Write each side of the equation inside a log.

5 𝑥 = 2 𝑥+2 becomes log(5 𝑥 ) = log(2 𝑥+2 ).

Step 2. Bring down the exponent in front of the logs

Use the log law log(a b ) = blog(a) to bring the powers down in front of the logs on both sides.

log(5 𝑥 ) = log(2 𝑥+2 ) becomes 𝑥log(5) = (𝑥+2)log2.

step 2 of solving an exponential equation

Step 3. Expand and collect 𝑥 terms

Expanding the bracket on the right hand side of the equation, (𝑥+2)log2 becomes 𝑥log2 +2log2.

Now the bracket is expanded, the equation becomes 𝑥log5 = 𝑥log2 + 2log.

We need to gather the 𝑥 terms all together on one side of the equation. We subtract 𝑥log2 from both sides of the equation which gives us:

𝑥log5 – 𝑥log2 = 2log2.

Step 4. Factorise and solve for 𝑥

Now the 𝑥 terms are on one side of the equation, we can factorise 𝑥 out.

𝑥log5 – 𝑥log2 becomes 𝑥(log5 – log2).

Therefore the equation becomes 𝑥(log5-log2) = 2log2.

To solve for 𝑥, we simply divide both sides of the equation by (log5-log2).

$x equals the fraction with numerator 2 log 2 and denominator open paren log 5 minus log 2 close paren$

This is the exact answer but it can be evaluated on a calculator as 𝑥 ≈ 1.51.

step 4 of solving an exponential equation

The image below shows a summary of how to solve the exponential equation in steps.

example of how to solve an exponential equation

Solving Exponential Equations with e

how to solve an exponential equation with e

The natural logarithm, ln(x) is the inverse function to e x .

Therefore ln(e x ) = x.

In this example, e 2x = 5.

Taking the natural logarithm of both sides, we get:

ln(e 2x ) = ln(5)

Here, ln and e cancel out so that:

Now we solve the resulting equation for x by dividing both sides by 2.

x = ln(5)/2

This is the exact answer and evaluating it on a calculator, we get x ≈ 0.805.

Here is another example.

Solve the exponential equation 5e 3x = 31.

When solving exponential equations with e, the natural logarithm ln(x) is used.

However it is important to first remove any coefficients from in front of the e.

We first divide both sides by 5 to get:

example of solving an exponential equation with e

Now the coefficient has been removed, the natural logarithm can be taken.

3x = ln(6.2)

Now we can solve for x by dividing both sides by 3.

x = ln(6.2)/3

Evaluating this on a calculator 𝑥 ≈ 0.608.

How to Solve Exponential Equation with the Same Base

If an exponential equation can be written so that both bases are the same, the equation can be solved by comparing the exponents. For example, 2 x+4 =8 x can be written as 2 x+4 =(2 3 ) x . Expanding, this becomes 2 x+4 =2 3x . Setting the exponents equal to each other, x+4=3x and so x = 2.

Solving Exponential Equations with the Same Base: Example 1

In this example, 8 is a power of 2 and so, it is possible to write both sides of the equation with the same base of 2.

how to solve an exponential with the same base

Solving Exponential Equations with the Same Base: Example 2

Solve 5 2x = 5 1-x

In this example, the bases are already equal. The bases are both 5.

Therefore the exponents must be equal to each other.

2x = 1 – x

Solve this equation for x by adding x to both sides.

Solve for x by dividing both sides of the equation by 3.

Solving Exponential Equations with the Same Base: Example 3

9 raised to the 2 x minus 1 power equals the square root of 3

Firstly, write both sides of the equation with the same base.

We can use 9 = 3 2 and √3 = 3 1 / 2 to help us.

3 squared raised to the exponent 2 x minus 1 end exponent equals 3 raised to the one half power

Now expanding, this becomes:

3 raised to the 4 x minus 2 power equals 3 raised to the one half power

Now the bases are equal, the exponents must be equal.

4𝑥 – 2 = 0.5

Now we solve for 𝑥. We add 2 to both sides:

Now we divide by 4 to get:

How to Solve Exponential Equations with Fractional Bases

Solving an Exponential Equation with a Fractional Base: Example 1

1 lines Line 1: one fourth to the x-th power equals 8

Step 1. Find a common base

Comparing both sides of the equation, 4 and 8 are both powers of 2.

one fourth to the x-th power equals 2 to the negative 2 power to the x-th power

Step 2. Equate the exponents

Now that the bases are equal, the exponents can be equated.

This can be solved for 𝑥 by dividing both sides by -2.

Solving an Exponential Equation with a Fractional Base: Example 2

1 lines Line 1: two thirds raised to the x plus 6 power equals open paren 8 over 27 close paren raised to the 3 x power

In this example, we can compare the fractions on each base.

8 = 2 3 and 27 = 3 3 .

open paren 8 over 27 close paren equals two thirds cubed

Now that the bases are the same, the exponents can be equated.

This equation can be solved by first subtracting 𝑥 from both sides to get 6 = 8𝑥.

Solving an Exponential Equation with a Fractional Base: Example 3

For exponential equations involving fractions, where no common base can be found, take logarithms of both sides. Then bring down the power in front of the log and solve for 𝑥 .

four fifths raised to the 2 x power equals 12

Step 3. Solve for 𝑥

Evaluating this on a calculator, 𝑥 = -5.57.

How to Solve Exponential Equations Involving Quadratics

An exponential equation will lead to a quadratic equation if one base is the square of the other base. For example the equation 9 x -5(3 x )+6=0 can be written as (3 x ) 2 -5(3 x )+6=0 . Substituting k=3 x , this leads to the quadratic equation k 2 -5k+6=0. This quadratic equation can be solved to find x.

The steps for solving an exponential equation involving a quadratic are:

Solving an Exponential Equation Leading to a Quadratic: Example 1

9 to the x-th power minus 5 times open paren 3 to the x-th power close paren plus 6 equals 0

We know that this equation will lead to a quadratic because 9 x is the square of 3 x .

Step 1. Write one base as the other base squared

9 x = (3 2 ) x which can be written as (3 x ) 2 .

Therefore 9 x = (3 x ) 2 .

3 to the x-th power squared minus 5 times open paren 3 to the x-th power close paren plus 6 equals 0

Step 2. Substitute this base for k to form a quadratic

Step 3. Solve the resulting quadratic for k

Therefore k = 3 or k = 2.

Step 4. Solve the exponential equations for x

Since we set k = 3 x , if k = 3 or k = 2 then 3 x = 3 or 3 x = 2.

If 3 x = 3 then x = 1.

If 3 x = 2 then we can solve this exponential equation with logs.

log(3 x )=log(2) and so, xlog(3) = log(2).

Solving an Exponential Equation Leading to a Quadratic: Example 2

25 to the x-th power minus 5 raised to the x plus 1 power plus 4 equals 0

Step 1. Write one base as the other squared

25 x = (5 x ) 2

5 to the x-th power squared minus 5 raised to the x plus 1 power plus 4 equals 0

Therefore k = 4 or k = 1.

Since k = 5 x , the solutions of k = 4 and k = 1 become 5 x = 4 and 5 x = 1.

If 5 x = 1, 𝑥 = 0

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Solve exponential equations using exponent properties

Solving Exponential Equations with Logarithms

From the Definition With Logarithms With Calculators

Most exponential equations do not solve neatly; there will be no way to convert the bases to being the same, such as the conversion of 4 and 8 into powers of 2 . In solving these more-complicated equations, you will have to use logarithms.

Taking logarithms will allow us to take advantage of the log rule that says that powers inside a log can be moved out in front as multipliers. By taking the log of an exponential, we can then move the variable (being in the exponent that's now inside a log) out in front, as a multiplier on the log. In other words, the log rule will let us move the variable back down onto the ground, where we can get our hands on it.

For instance:

Content Continues Below

Solve 2 x = 30

If this equation had asked me to "Solve 2 x = 32 ", then finding the solution would have been easy, because I could have converted the 32 to 2 5 , set the exponents equal, and solved for " x = 5 ". But, unlike 32 , 30 is not a power of 2 so I can't set powers equal to each other. I need some other method of getting at the x , because I can't solve with the equation with the variable floating up there above the 2 ; I need it back down on the ground where it belongs, where I can get at it. And I'll have to use logarithms to bring that variable down.

When dealing with equations, I can do whatever I like to the equation, as long as I do the same thing to both sides. And, to solve an equation, I have to get the variable by itself on one side of the "equals" sign; to isolate the variable, I have to "undo" whatever has been done to the variable.

In this case, the variable x has been put in the exponent. The backwards (technically, the " inverse ") of exponentials are logarithms , so I'll need to undo the exponent by taking the log of both sides of the equation. This is useful to me because of the log rule that says that exponents inside a log can be turned into multipliers in front of the log:

log b ( m n ) = n · log b ( m )

When I take the log of both sides of an equation, I can use any log I like (base- 10 log, base- 2 log, natural log, etc), but some are sometimes more useful than others. Since the base in the equation " 2 x = 30 " is " 2 ", I might try using a base- 2 log:

log 2 (2 x ) = log 2 (30)

Any log of the log's base returns a value of 1 , so log 2 (2) = 1 . Then:

x · log 2 (2) = log 2 (30)

x (1) = log 2 (30)

x = log 2 (30)

But we can't evaluate this expression in our calculators as it stands. First, we'd need to apply the change-of-base formula to convert the expression into something in a base that our calculators can understand; namely, the natural log or the common log. That conversion looks like this:

Reminder: The " ln " is the abbreviation for "logarithmus naturalis", the Latin version of what became "natural log" in English. The abbreviation is pronounced "ell-enn" and written with a lower-case "L" followed by a lower-case "N". There is no "I" ("eye") in the function name!

What would happen if I just used the natural log, instead of a base-two log, in the first place? The process would have been exactly the same, and the eventual answer would have been equivalent.

ln(2 x ) = ln(30)

x · ln(2) = ln(30)

Either way, I get the same answer, but taking natural log in the first place was simpler and shorter.

Note: I could have used the common (base- 10 ) log instead of the natural (that is, the base- e ) log, and still come up with the same value (when evaluated in the calculator).

Since science uses the natural log so much, and since it is one of the two logs that calculators can evaluate, I tend to take the natural log of both sides when solving exponential equations. This is not (generally) required, but is often more useful than other options.

Solve 5 x = 212 . Give your answer in exact form and as a decimal approximation to three places.

Since 212 is not a power of 5 , then I will have to use logs to solve this equation. I could take base- 5 log of each side, solve, and then apply the change-of-base formula, but I think I'd rather just use the natural log in the first place:

ln(5 x ) = ln(212)

x · ln(5) = ln(212)

...or about 3.328 , rounded to three decimal places.

Solve 10 2 x = 52

Since 52 is not a power of 10 , I will have to use logs to solve this. In this particular instance, since the base is 10 and since base- 10 logs can be done on the calculator, I will use the common log instead of the natural log to solve this particular equation:

10 2 x = 52

log(10 2 x ) = log(52)

2 x · log (10) = log(52)

2 x (1) = log(52)

2 x = log(52)

...or about 0.858 , rounded to three decimal places.

Solve 3(2 x +4 ) = 350

Before I can start looking at the exponential, I first have to get rid of the 3 , so I'll divide that off to get:

...or about 2.866 , rounded to three decimal places.

Note: You could also solve the above by using exponent rules to break apart the power on the 2 :

2 x +4 = (2 x )(2 4 ) = (2 x )(16)

Then divide through by the 16 and simplify to get:

Then take the log of each side. You'll get an answer in the form:

When you evaluate this, you'll get the same decimal equivalent, 2.866 , in your calculator. Don't be shy about being flexible!

URL: https://www.purplemath.com/modules/simpexpo2.htm

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Exponential Equations

1. Solve in real numbers:

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27. Solve in real numbers:

28. Solve in real numbers:

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Module 12: Exponential and Logarithmic Equations and Models

Exponential equations, learning outcomes.

The first technique we will introduce for solving exponential equations 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 [latex]b>0,\text{ }b\ne 1[/latex], [latex]{b}^{S}={b}^{T}[/latex] 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 [latex]{3}^{4x - 7}=\frac{{3}^{2x}}{3}[/latex]. 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 :

[latex]\begin{array}{l}{3}^{4x - 7}\hfill & =\frac{{3}^{2x}}{3}\hfill & \hfill \\ {3}^{4x - 7}\hfill & =\frac{{3}^{2x}}{{3}^{1}}\hfill & {\text{Rewrite 3 as 3}}^{1}.\hfill \\ {3}^{4x - 7}\hfill & ={3}^{2x - 1}\hfill & \text{Use the division property of exponents}\text{.}\hfill \\ 4x - 7\hfill & =2x - 1\text{ }\hfill & \text{Apply the one-to-one property of exponents}\text{.}\hfill \\ 2x\hfill & =6\hfill & \text{Subtract 2}x\text{ and add 7 to both sides}\text{.}\hfill \\ x\hfill & =3\hfill & \text{Divide by 3}\text{.}\hfill \end{array}[/latex]

A General Note: 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 [latex]b\ne 1[/latex],

[latex]{b}^{S}={b}^{T}\text{ if and only if }S=T[/latex]

How To: Given an exponential equation Of the form [latex]{b}^{S}={b}^{T}[/latex], where S and T are algebraic expressions with an unknown, solve for the unknown

Example: Solving an Exponential Equation with a Common Base

Solve [latex]{2}^{x - 1}={2}^{2x - 4}[/latex].

[latex]\begin{array}{l} {2}^{x - 1}={2}^{2x - 4}\hfill & \text{The common base is }2.\hfill \\ \text{ }x - 1=2x - 4\hfill & \text{By the one-to-one property the exponents must be equal}.\hfill \\ \text{ }x=3\hfill & \text{Solve for }x.\hfill \end{array}[/latex]

Solve [latex]{5}^{2x}={5}^{3x+2}[/latex].

[latex]x=–2[/latex]

Rewriting 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 [latex]256={4}^{x - 5}[/latex]. 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 :

[latex]\begin{array}{l}256={4}^{x - 5}\hfill & \hfill \\ {2}^{8}={\left({2}^{2}\right)}^{x - 5}\hfill & \text{Rewrite each side as a power with base 2}.\hfill \\ {2}^{8}={2}^{2x - 10}\hfill & \text{To take a power of a power, multiply the exponents}.\hfill \\ 8=2x - 10\hfill & \text{Apply the one-to-one property of exponents}.\hfill \\ 18=2x\hfill & \text{Add 10 to both sides}.\hfill \\ x=9\hfill & \text{Divide by 2}.\hfill \end{array}[/latex]

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: Solving Equations by Rewriting Them to Have a Common Base

Solve [latex]{8}^{x+2}={16}^{x+1}[/latex].

[latex]\begin{array}{llllll}\text{ }{8}^{x+2}={16}^{x+1}\hfill & \hfill \\ {\left({2}^{3}\right)}^{x+2}={\left({2}^{4}\right)}^{x+1}\hfill & \text{Write }8\text{ and }16\text{ as powers of }2.\hfill \\ \text{ }{2}^{3x+6}={2}^{4x+4}\hfill & \text{To take a power of a power, multiply the exponents}.\hfill \\ \text{ }3x+6=4x+4\hfill & \text{Use the one-to-one property to set the exponents equal to each other}.\hfill \\ \text{ }x=2\hfill & \text{Solve for }x.\hfill \end{array}[/latex]

Solve [latex]{5}^{2x}={25}^{3x+2}[/latex].

[latex]x=–1[/latex]

Example: Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base

Solve [latex]{2}^{5x}=\sqrt{2}[/latex].

[latex]\begin{array}{l}{2}^{5x}={2}^{\frac{1}{2}}\hfill & \text{Write the square root of 2 as a power of }2.\hfill \\ 5x=\frac{1}{2}\hfill & \text{Use the one-to-one property}.\hfill \\ x=\frac{1}{10}\hfill & \text{Solve for }x.\hfill \end{array}[/latex]

Solve [latex]{5}^{x}=\sqrt{5}[/latex].

[latex]x=\frac{1}{2}[/latex]

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: Determine When an Equation has No Solution

Solve [latex]{3}^{x+1}=-2[/latex].

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.

Analysis of the Solution

The figure below shows that the two graphs do not cross so the left side of the equation is never equal to the right side of the equation. Thus the equation has no solution.

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

Solve [latex]{2}^{x}=-100[/latex].

The equation has no solution.

Using Logarithms to Solve Exponential Equations

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 that since [latex]\mathrm{log}\left(a\right)=\mathrm{log}\left(b\right)[/latex] is equal to a = b , we may apply logarithms with the same base to both sides of an exponential equation.

How To: Given an exponential equation Where a common base cannot be found, solve for the unknown

Example: Solving an Equation Containing Powers of Different Bases

Solve [latex]{5}^{x+2}={4}^{x}[/latex].

[latex]\begin{array}{l}\text{ }{5}^{x+2}={4}^{x}\hfill & \text{There is no easy way to get the powers to have the same base}.\hfill \\ \text{ }\mathrm{ln}{5}^{x+2}=\mathrm{ln}{4}^{x}\hfill & \text{Take ln of both sides}.\hfill \\ \text{ }\left(x+2\right)\mathrm{ln}5=x\mathrm{ln}4\hfill & \text{Use the power rule for logs}.\hfill \\ \text{ }x\mathrm{ln}5+2\mathrm{ln}5=x\mathrm{ln}4\hfill & \text{Use the distributive property}.\hfill \\ \text{ }x\mathrm{ln}5-x\mathrm{ln}4=-2\mathrm{ln}5\hfill & \text{Get terms containing }x\text{ on one side, terms without }x\text{ on the other}.\hfill \\ x\left(\mathrm{ln}5-\mathrm{ln}4\right)=-2\mathrm{ln}5\hfill & \text{On the left hand side, factor out }x.\hfill \\ \text{ }x\mathrm{ln}\left(\frac{5}{4}\right)=\mathrm{ln}\left(\frac{1}{25}\right)\hfill & \text{Use the properties of logs}.\hfill \\ \text{ }x=\frac{\mathrm{ln}\left(\frac{1}{25}\right)}{\mathrm{ln}\left(\frac{5}{4}\right)}\hfill & \text{Divide by the coefficient of }x.\hfill \end{array}[/latex]

Solve [latex]{2}^{x}={3}^{x+1}[/latex].

[latex]x=\frac{\mathrm{ln}3}{\mathrm{ln}\left(\frac{2}{3}\right)}[/latex]

Is there any way to solve [latex]{2}^{x}={3}^{x}[/latex]?

Yes. The solution is x = 0.

Equations Containing [latex]e[/latex]

One common type of exponential equations are those with base e . This constant occurs again and again in nature, mathematics, science, engineering, and 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 [latex]y=A{e}^{kt}[/latex], solve for [latex]t[/latex]

TIPS for Success

Just as you have done when solving various types of equations, isolate the term containing the variable for which you are solving before applying any properties of equality or inverse operations. That’s why, in the example above, you must divide away the [latex]A[/latex] first.

Remember that the functions [latex]y=e^{x}[/latex] and [latex]y=\mathrm{ln}\left(x\right)[/latex] are inverse functions.

Therefore, [latex]\mathrm{ln}\left({e}^{x}\right)=x[/latex] for all [latex]x[/latex] and [latex]e^{\mathrm{ln}\left(x\right)}=x[/latex] for [latex]x>0[/latex].

Example: Solve an Equation of the Form [latex]y=A{e}^{kt}[/latex]

Solve [latex]100=20{e}^{2t}[/latex].

[latex]\begin{array}{l}100\hfill & =20{e}^{2t}\hfill & \hfill \\ 5\hfill & ={e}^{2t}\hfill & \text{Divide by the coefficient 20}\text{.}\hfill \\ \mathrm{ln}5\hfill & =\mathrm{ln}{{e}^{2t}}\hfill & \text{Take ln of both sides.}\hfill \\ \mathrm{ln}5\hfill & =2t\hfill & \text{Use the fact that }\mathrm{ln}\left(x\right)\text{ and }{e}^{x}\text{ are inverse functions}\text{.}\hfill \\ t\hfill & =\frac{\mathrm{ln}5}{2}\hfill & \text{Divide by the coefficient of }t\text{.}\hfill \end{array}[/latex]

Using laws of logs, we can also write this answer in the form [latex]t=\mathrm{ln}\sqrt{5}[/latex]. If we want a decimal approximation of the answer, then we use a calculator.

Solve [latex]3{e}^{0.5t}=11[/latex].

[latex]t=2\mathrm{ln}\left(\frac{11}{3}\right)[/latex] or [latex]\mathrm{ln}{\left(\frac{11}{3}\right)}^{2}[/latex]

Does every equation of the form [latex]y=A{e}^{kt}[/latex] have a solution?

No. There is a solution when [latex]k\ne 0[/latex], and when [latex]y[/latex] and [latex] A [/latex] 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 [latex]2=-3{e}^{t}[/latex].

Example: Solving an Equation That Can Be Simplified to the Form [latex]y=A{e}^{kt}[/latex]

Solve [latex]4{e}^{2x}+5=12[/latex].

[latex]\begin{array}{l}4{e}^{2x}+5=12\hfill & \hfill \\ 4{e}^{2x}=7\hfill & \text{Subtract 5 from both sides}.\hfill \\ {e}^{2x}=\frac{7}{4}\hfill & \text{Divide both sides by 4}.\hfill \\ 2x=\mathrm{ln}\left(\frac{7}{4}\right)\hfill & \text{Take ln of both sides}.\hfill \\ x=\frac{1}{2}\mathrm{ln}\left(\frac{7}{4}\right)\hfill & \text{Solve for }x.\hfill \end{array}[/latex]

Solve [latex]3+{e}^{2t}=7{e}^{2t}[/latex].

[latex]t=\mathrm{ln}\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{2}\mathrm{ln}\left(2\right)[/latex]

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 taking the logarithm of 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: Solving Exponential Functions in Quadratic Form

Solve [latex]{e}^{2x}-{e}^{x}=56[/latex].

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

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 [latex]{e}^{x}=-7[/latex] because a positive number never equals a negative number. The solution [latex]x=\mathrm{ln}\left(-7\right)[/latex] is not a real number and in the real number system, this solution is rejected as an extraneous solution.

Solve [latex]{e}^{2x}={e}^{x}+2[/latex].

[latex]x=\mathrm{ln}2[/latex]

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.

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Exponential Equations

Exponential equations , as the name suggests, involve exponents. We know that the exponent of a number (base) indicates the number of times the number (base) is multiplied. But, what happens if the power of a number is a variable? When the power is a variable and if it is a part of an equation, then it is called an exponential equation. We may need to use the connection between the exponents and logarithms to solve the exponential equations.

Let us learn the definition of exponential equations along with the process of solving them when the bases are the same and when the bases are not the same along with a few solved examples and practice questions.

What are Exponential Equations?

An exponential equation is an equation with exponents where the exponent (or) a part of the exponent is a variable . For example, 3 x = 81, 5 x - 3 = 625, 6 2y - 7 = 121, etc are some examples of exponential equations. We may come across the use of exponential equations when we are solving the problems of algebra, compound interest , exponential growth , exponential decay , etc.

Types of Exponential Equations

There are three types of exponential equations. They are as follows:

Equations with the same bases on both sides. (Example: 4 x = 4 2 )
Equations with different bases that can be made the same. (Example: 4 x = 16 which can be written as 4 x = 4 2 )
Equations with different bases that cannot be made the same. (Example: 4 x = 15)

Equations with Exponents

The equations in algebra involving variable exponents are called equations with exponents or exponential equations. In other words, we can say that algebraic equations in which variables occur as exponents are known as the equations with exponents. Some of the examples of such an equation are, 3 x + 4 = 81, -2 3y-7 = -64, etc.

Exponential Equations Formulas

While solving an exponential equation , the bases on both sides may be the same or may not be the same. Here are the formulas that are used in each of these cases, which we will learn in detail in the upcoming sections.

Property of Equality for Exponential Equations

This property is useful to solve an exponential equation with the same bases. It says when the bases on both sides of an exponential equation are equal, then the exponents must also be equal. i.e.,

a x = a y ⇔ x = y.

Exponential Equations to Logarithmic Form

We know that logarithms are nothing but exponents and vice versa. Hence an exponential equation can be converted into a logarithmic function . This helps in the process of solving an exponential equation with different bases. Here is the formula to convert exponential equations into logarithmic equations.

b x = a ⇔ log b a = x

Solving Exponential Equations With Same Bases

Sometimes, an exponential equation may have the same bases on both sides of the equation. For example, 5 x = 5 3 has the same base 5 on both sides. Sometimes, though the exponents on both sides are not the same, they can be made the same. For example, 5 x = 125. Though it doesn't have the same bases on both sides of the equation, they can be made the same by writing it as 5 x = 5 3 (as 125 = 5 3 ). To solve the exponential equations in each of these cases, we just apply the property of equality of exponential equations, using which, we set the exponents to be the same and solve for the variable.

Here is another example where the bases are not the same but can be made the same.

Example: Solve the exponential equation 7 y + 1 = 343 y .

We know that 343 = 7 3 . Using this, the given equation can be written as,

7 y + 1 = (7 3 ) y

7 y + 1 = 7 3y

Now the bases on both sides are the same. So we can set the exponents to be the same.

Subtracting y from both sides,

Dividing both sides by 2,

Solving Exponential Equations With Different Bases

Sometimes, the bases on both sides of an exponential equation may not be the same (or) cannot be made the same. We solve the exponential equations using logarithms when the bases are not the same on both sides of the equation. For example, 5 x = 3 neither has the same bases on both sides nor the bases can be made the same. In such cases, we can do one of the following things.

Convert the exponential equation into the logarithmic form using the formula b x = a ⇔ log b a = x and solve for the variable.
Apply logarithm (log) on both sides of the equation and solve for the variable. In this case, we will have to use a property of logarithm , log a m = m log a.

We will solve the equation 5 x = 3 in each of these methods.

We will convert 5 x = 3 into logarithmic form. Then we get,

log 5 3 = x

Using the change of base property ,

x = (log 3) / (log 5)

We will apply log on both sides of 5 x = 3. Then we get, log 5 x = log 3. Using the property log a m = m log a on the left side of the equation, we get, x log 5 = log 3. Dividing both sides by log 5,

Important Notes on Exponential Equations:

Here are some important notes with respect to the exponential equations.

To solve the exponential equations of the same bases, just set the exponents equal.
To solve the exponential equations of different bases, apply logarithm on both sides.
The exponential equations with the same bases also can be solved using logarithms.
If an exponential equation has 1 on any one side, then we can write it as 1 = a 0 , for any 'a'. For example, to solve 5 x = 1, we can write it as 5 x = 5 0 , then we get x = 0.
To solve an exponential equation using logarithms, we can either apply "log" or apply "ln" on both sides.

Exponential Form
Exponent Rules
Exponential Functions
Exponential Equation Calculator

Exponential Equations Examples

Example 1: Solve 27 / (3 -x ) = 3 6 .

We know that 27 = 3 3 . We can make the bases to be the same on both sides using this.

3 3 / (3 -x ) = 3 6

Using the quotient property of exponents, a m /a n = a m - n . Using this,

3 3 - (-x) = 3 6

3 3 + x = 3 6

Now the bases on both sides are the same. So we can set the exponents to be equal.

Subtracting 3 from both sides,

Therefore, the solution of the given exponential equation is x = 3.

Example 2: Solve the exponential equation 7 3x + 7 = 490.

490 cannot be written as a power of 7. So we cannot make the bases to be the same here. So we solve this exponential equation using logarithms.

Apply log on both sides of the given equation,

log 7 3x + 7 = log 490

Using a property of logarithms , log a m = m log a. Using this,

(3x + 7) log 7 = log 490 ... (1)

Here, 490 = 49 × 10 = 7 2 × 10.

So, log 490 = log (7 2 × 10)

= log 7 2 + log 10 (because log (mn) = log m + log n)

= 2 log 7 + 1 (because log a m = m log a and log 10 = 1)

Substituting this in (1),

(3x + 7) log 7 = 2 log 7 + 1

Dividing both sides by log 7,

3x + 7 = (2 log 7 + 1) / (log 7)

3x + 7 = 2 + (1 / log 7)

Subtracting 7 from both sides,

3x = -5 + (1 / log 7)

Dividing both sides by 3,

x = -5/3 + (1 / (3 log 7))

Answer: The solution of the given exponential equation is x = -5/3 + (1 / (3 log 7)).

Example 3: How long does it take for $20,000 to double if the amount is compounded annually at 8% annual interest? Round your answer to the nearest integer.

The principal amount, P = $20000.

The rate of interest is, r = 8% = 8/100 = 0.08.

The final amount is, A = 20000 x 2 = $40,000

Let us assume that the required time in years is t.

Using the compound interest formula when compounded annually,

A = P (1 + r) t

40000 = 20000 (1 + 0.08) t

Dividing both sides by 20000,

2 = (1.08) t

Taking log on both sides,

log 2 = log (1.08) t

log 2 = t log (1.08)

t = (log 2) / (log 1.08)

The final answer is rounded to the nearest integer.

Answer: It takes 9 years for $20,000 to double itself.

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Exponential Equations Questions

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FAQs on Exponential Equations

What are exponential equations.

An exponential equation is an equation that has a variable in its exponent(s). For example, 5 2x - 3 = 125, 3 7 - 2x = 91, etc are exponential equations.

What Are Types of Exponential Equations?

There are three types of exponential equations. They are,

The exponential equations with the same bases on both sides.
The exponential equations with different bases on both sides that can be made the same.
The exponential equations with different bases on both sides that cannot be made the same.

How To Solve Exponential Equations?

To solve the exponential equations of equal bases, we set the exponents equal whereas to solve the exponential equations of different bases, we apply logarithms on both sides.

How To Write Exponential Equation in Logarithmic Form?

Writing the exponential equation in the logarithmic form helps us to solve it. This can be done using the formula b x = a ⇔ log b a = x.

What Is the Property of Equality of Exponential Equations?

The equality property of exponential equations says to set the exponents equal whenever the bases on both sides of the equation are equal. i.e., a x = a y ⇔ x = y.

How To Solve Exponential Equations of Same bases?

When an exponential equation has the same bases on both sides, just set the exponents equal and solve for the variable. Here is an example, 4 2x - 1 = 4 1 - x . Here the bases on both sides are equal. So we can set the exponents equal. 2x - 1 = 1 - x 3x = 2 x = 2/3.

How To Solve Exponential Equations of Different bases?

When an exponential equation has different bases on both sides, apply log on both sides and solve for the variable. Here is an example, 4 x - 5 = 8. Taking log on both sides, log 4 x - 5 = log 8 (x - 5) log 4 = log 8 x - 5 = (log 8) / (log 4) x = [(log 8) / (log 4)] + 5.

How To Solve Exponential Equations Using Logarithms?

We solve exponential equations using logarithms in two ways.

Convert the exponential equation into logarithmic equation using b x = a ⇔ log b a = x.
Apply "log" or "ln" on both sides and solve.

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