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Solving Exponential Equations Using Other Methods

Author: Sophia
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what's covered
In this lesson, you will solve exponential equations by using logarithms, which is necessary when both sides can’t be written in terms of the same base. Specifically, this lesson will cover:

Table of Contents

1. Solving Equations Involving One Exponential Function

When the bases can’t be equated, logarithms are required. Through the one-to-one property of logarithms, we know that for positive quantities R and S, if R equals S comma then log subscript b R equals log subscript b S.

hint
Assume that both sides of the equation represent positive quantities.
  • When the exponential function has base 10, apply the common logarithm to both sides.
  • When the exponential function has base e, apply the natural logarithm to both sides.
  • For all other bases, either logarithm is equally as effective.

EXAMPLE

Solve the equation open parentheses 1.02 close parentheses to the power of 4 t end exponent equals 1.5.

It seems that 1.5 is not a power of 1.02. Since both quantities are positive, apply a logarithm to both sides. In this example, the common logarithm will be used.

open parentheses 1.02 close parentheses to the power of 4 t end exponent equals 1.5 This is the original equation.
log   open parentheses 1.02 close parentheses to the power of 4 t end exponent equals log     1.5 Apply the common logarithm to both sides.
4 t times log   open parentheses 1.02 close parentheses equals log     1.5 Use the power property of logarithms, log subscript b open parentheses x to the power of y close parentheses equals y times log subscript b x.
t equals fraction numerator log   1.5 over denominator 4   log   open parentheses 1.02 close parentheses end fraction Divide both sides by 4   log   open parentheses 1.02 close parentheses.

The exact solution is t equals fraction numerator log   1.5 over denominator 4   log   open parentheses 1.02 close parentheses end fraction. Rounded to three decimal places, t almost equal to 5.119.

Note: the natural logarithm could have also been applied to both sides.

In this case, t equals fraction numerator ln   1.5 over denominator 4   ln   open parentheses 1.02 close parentheses end fraction comma which has the same exact value as t equals fraction numerator log   1.5 over denominator 4   log   open parentheses 1.02 close parentheses end fraction.

In general, here is the sequence of steps that are required to solve an exponential equation by using logarithms.

step by step
  1. Isolate the exponential term on one side of the equation.
  2. Check to make sure that the quantities on both sides of the equation are positive.
  3. Apply the appropriate logarithm to both sides of the equation, then simplify.
  4. Solve for the variable.

EXAMPLE

Solve the equation 50 e to the power of short dash 0.2 t end exponent plus 72 equals 85.

The first goal is to isolate the exponential term on one side.

Next, divide both sides by the coefficient of the exponential expression.

Then, apply logarithms and solve for t. Since this equation contains base e, the natural logarithm will be used.

50 e to the power of short dash 0.2 t end exponent plus 72 equals 85 This is the original equation.
50 e to the power of short dash 0.2 t end exponent equals 13 Isolate the exponential term on one side. Subtract 72 from both sides.
e to the power of short dash 0.2 t end exponent equals 13 over 50
e to the power of short dash 0.2 t end exponent equals 0.26 		Divide both sides by 50. Since 13 over 50 is a terminating decimal, writing it in decimal form doesn’t cause a loss in accuracy.
ln     e to the power of short dash 0.2 t end exponent equals ln     0.26 Apply the natural logarithm to both sides.
short dash 0.2 t equals ln     0.26 Use the inverse property ln     e to the power of k equals k.
t equals fraction numerator ln   0.26 over denominator short dash 0.2 end fraction Divide both sides by -0.2.

The final answer is t equals fraction numerator ln   0.26 over denominator short dash 0.2 end fraction comma which can also be written t equals short dash fraction numerator ln   0.26 over denominator 0.2 end fraction. Rounded to three decimal places, t almost equal to 6.735.

try it
Consider the equation 5 times 3 to the power of short dash 0.4 t end exponent plus 12 equals 14.
Solve the equation for t, giving the exact solution.
To solve for t, first isolate the exponential expression to one side, then use logarithms as necessary.

5 times 3 to the power of short dash 0.4 t end exponent plus 12 equals 14 This is the original equation.
5 times 3 to the power of short dash 0.4 t end exponent equals 2 Subtract 12 from both sides.
3 to the power of short dash 0.4 t end exponent equals 0.4 Divide both sides by 5. Write 2 over 5 in decimal form to make the expression look less complicated.
log   3 to the power of short dash 0.4 t end exponent equals log   0.4 Apply the “log” to both sides to bring the exponent down.
short dash 0.4 t times log   3 equals log   0.4 Use the power property of logarithms to bring the exponent down.
t equals fraction numerator log   0.4 over denominator short dash 0.4   log   3 end fraction Divide both sides by short dash 0.4 log   3. This is the solution in exact form.

The exact form of the solution is t equals fraction numerator log   0.4 over denominator short dash 0.4   log   3 end fraction.
Solve the equation for t, giving the approximate solution rounded to three decimal places.
t almost equal to 2.085

Let’s look at an example with a more complicated exponent.

EXAMPLE

Consider the equation 10 to the power of x minus 3 end exponent equals 21.

Since the exponential function has base 10, we will apply the common logarithm to both sides.

10 to the power of x minus 3 end exponent equals 21 This is the original equation.
log   open parentheses 10 to the power of x minus 3 end exponent close parentheses equals log     21 Apply the common logarithm to both sides.
x minus 3 equals log     21 Use the inverse property log subscript b open parentheses b to the power of x close parentheses equals x.
x equals 3 plus log     21 Add 3 to both sides.

The exact solution is x equals 3 plus log     21. This could also be written as x equals log     21 plus 3 comma but this could look ambiguous, and one might mistakenly say this is the same as log     24.

Rounded to three decimal places, x almost equal to 4.322.

try it
Consider the equation e to the power of 2 x minus 1 end exponent equals 15.
Solve the equation for x, giving the exact solution.
The exponential term is isolated to one side. Since the base is e, apply the natural logarithm to both sides, then solve for x.

e to the power of 2 x minus 1 end exponent equals 15 This is the original equation.
ln   e to the power of 2 x minus 1 end exponent equals ln   15 Apply the natural logarithm to both sides.
2 x minus 1 equals ln   15 Use the property ln   e to the power of k equals k.
2 x equals 1 plus ln   15 Add 1 to both sides.
x equals fraction numerator 1 plus ln   15 over denominator 2 end fraction Divide both sides by 2. This is the exact form of the solution.

The exact form of the solution is x equals fraction numerator 1 plus ln   15 over denominator 2 end fraction.
Solve the equation for x, giving the approximate solution rounded to three decimal places.
x almost equal to 1.854

Let’s also explore an exponential equation whose base is neither 10 nor e.

EXAMPLE

Consider the equation 4 to the power of x plus 2 end exponent equals 35.

Since 35 cannot be written as a power of 4, apply a logarithm to both sides. In this example, we’ll use the common logarithm.

4 to the power of x plus 2 end exponent equals 35 This is the original equation.
log   open parentheses 4 to the power of x plus 2 end exponent close parentheses equals log     35 Apply the common logarithm to both sides.
Note: each side is positive; therefore, both logarithms are defined.
open parentheses x plus 2 close parentheses   log     4 equals log     35 Use the power property of logarithms, log subscript b open parentheses x to the power of y close parentheses equals y times log subscript b x.
x   log     4 plus 2   log     4 equals log     35 Distribute log     4 on the left side.
x   log     4 equals log     35 minus 2   log     4 Subtract 2   log     4 from both sides.
x equals fraction numerator log     35 minus 2   log     4 over denominator log     4 end fraction Divide both sides by log     4.

Thus, the answer exact is x equals fraction numerator log     35 minus 2   log     4 over denominator log     4 end fraction. When approximated to three decimal places, x almost equal to 0.565.

try it
Consider the equation 2 to the power of x minus 3 end exponent plus 8 equals 11.
Solve the equation for x, giving the exact solution.
To solve, isolate the exponential term to one side, then apply logarithms as necessary.

2 to the power of x minus 3 end exponent plus 8 equals 11 This is the original equation.
2 to the power of x minus 3 end exponent equals 3 Subtract 8 from both sides.
log   2 to the power of x minus 3 end exponent equals log   3 Since 3 is not a rational power of 2, apply the logarithm (base 10) to both sides.
open parentheses x minus 3 close parentheses   log   2 equals log   3 Use the power property of logarithms.
x   log   2 minus 3   log   2 equals log   3 Distribute.
x   log   2 equals log   3 plus 3   log   2 Add 3   log   2 to both sides.
x equals fraction numerator log   3 plus 3   log   2 over denominator log   2 end fraction Divide both sides by log   2. This is the exact form of the answer.

The exact form of the solution is x equals fraction numerator log   3 plus 3   log   2 over denominator log   2 end fraction. Note: If you apply “ln” to both sides instead of “log”, the solution x equals fraction numerator ln   3 plus 3   ln   2 over denominator ln   2 end fraction.
Solve the equation for x, giving the approximate solution rounded to three decimal places.
x almost equal to 4.585

2. Solving Equations Involving Two Exponential Functions

First, let’s look at exponential functions that are set equal to each other.

EXAMPLE

Solve the equation e to the power of x plus 1 end exponent equals 2 to the power of 4 minus x end exponent.

Since one exponential function has base e, apply the natural logarithm to both sides. Since there is an exponential function on both sides of the equation, each side of the equation is positive.

e to the power of x plus 1 end exponent equals 2 to the power of 4 minus x end exponent This is the original equation.
ln   open parentheses e to the power of x plus 1 end exponent close parentheses equals ln   open parentheses 2 to the power of 4 minus x end exponent close parentheses Apply the natural logarithm on both sides.
x plus 1 equals open parentheses 4 minus x close parentheses   ln     2 On the left side, use the inverse property ln   open parentheses e to the power of x close parentheses equals x.
On the right side, use the power property log subscript b open parentheses x to the power of y close parentheses equals y times log subscript b x.
x plus 1 equals 4   ln     2 minus x   ln     2 Distribute on the right side.
x plus 1 plus x   ln     2 equals 4   ln     2 Collect all terms with x on one side of the equation.
x plus x   ln     2 equals short dash 1 plus 4   ln     2 Subtract 1 from both sides.
x open parentheses 1 plus ln     2 close parentheses equals short dash 1 plus 4   ln     2 Factor out x on the left side.
x equals fraction numerator short dash 1 plus 4   ln     2 over denominator 1 plus ln     2 end fraction Divide both sides by 1 plus ln     2.

The solution in exact form is x equals fraction numerator short dash 1 plus 4   ln     2 over denominator 1 plus ln     2 end fraction. Rounded to three decimal places, x almost equal to 1.047.

watch
Here is the worked-out solution to 3 to the power of 2 x minus 1 end exponent equals 5 to the power of x plus 7 end exponent.

try it
Consider the equation e to the power of 2 x end exponent equals 5 to the power of x plus 2 end exponent.
Solve the equation for x, giving the exact solution.
Since there is a base e term, apply the natural logarithm to both sides, then solve.

e to the power of 2 x end exponent equals 5 to the power of x plus 2 end exponent This is the original equation.
ln   e to the power of 2 x end exponent equals ln   5 to the power of x plus 2 end exponent Apply the natural logarithm to both sides.
2 x equals open parentheses x plus 2 close parentheses   ln   5 On the left side, use the property ln   e to the power of k equals k.
On the right, use the power property to bring the exponent down.
2 x equals x   ln   5 plus 2   ln   5 Distribute on the right-hand side.
2 x minus x   ln   5 equals 2   ln   5 Subtract x   ln   5 from both sides to get all x terms to one side.
x open parentheses 2 minus ln   5 close parentheses equals 2   ln   5 Factor out x. This sets us up to solve for x.
x equals fraction numerator 2   ln   5 over denominator 2 minus ln   5 end fraction Divide both sides by 2 minus ln   5. This is the exact form of the solution.

The exact form of the solution is x equals fraction numerator 2   ln   5 over denominator 2 minus ln   5 end fraction.
Solve the equation for x, giving the approximate solution rounded to three decimal places.
x almost equal to 8.242

We can also solve exponential functions that are quadratic in form.

EXAMPLE

Solve the equation 4 to the power of x minus 3 open parentheses 2 to the power of x close parentheses minus 18 equals 0.

Since this equation contains more than two terms, one-to-one properties will not help. At first, it looks as if there are two different bases in this equation, but notice that 4 is a power of 2.

Then, write 4 to the power of x equals open parentheses 2 squared close parentheses to the power of x equals 2 to the power of 2 x end exponent equals open parentheses 2 to the power of x close parentheses squared.

4 to the power of x minus 3 open parentheses 2 to the power of x close parentheses minus 18 equals 0 This is the original equation.
open parentheses 2 to the power of x close parentheses squared minus 3 open parentheses 2 to the power of x close parentheses minus 18 equals 0 4 to the power of x equals open parentheses 2 to the power of x close parentheses squared
u squared minus 3 u minus 18 equals 0 Let u equals 2 to the power of x.
open parentheses u minus 6 close parentheses open parentheses u plus 3 close parentheses equals 0
u equals 6 space or space u equals short dash 3 		Solve the equation for u.
2 to the power of x equals 6 Replace u with 2 to the power of x in the equation u equals 6.
ln     2 to the power of x equals ln     6 Apply the natural logarithm to both sides.
x   ln     2 equals ln     6 Use the property log subscript b open parentheses x to the power of y close parentheses equals y times log subscript b x.
x equals fraction numerator ln     6 over denominator ln     2 end fraction Divide both sides by ln     2.
2 to the power of x equals short dash 3 Replace u with 2 to the power of x in the equation u equals short dash 3.
No solution An exponential function can never be equal to a negative number; therefore, this equation has no solution.

The solution is x equals fraction numerator ln     6 over denominator ln     2 end fraction. Substituting this value into the original equation verifies that it is a solution.

watch
To see the worked-out solution of 49 to the power of x minus 4 times 7 x equals 12 comma view the following video.

try it
Consider the equation e to the power of 2 x end exponent minus 4 e to the power of x minus 5 equals 0.
Solve the equation for x.
This is quadratic in form since one power of e is double the other.

Write e to the power of 2 x end exponent equals open parentheses e to the power of x close parentheses squared comma then the equation becomes open parentheses e to the power of x close parentheses squared minus 4 e to the power of x minus 5 equals 0.

To make the equation less complicated, let u equals e to the power of x.

u squared minus 4 u minus 5 equals 0 The equation after e to the power of x is replaced by u.
open parentheses u minus 5 close parentheses open parentheses u plus 1 close parentheses equals 0 Factor the quadratic.
u equals 5 comma space u equals short dash 1 Set each factor equal to 0 and solve.

For each value of u, find the corresponding value of x.

For u equals 5 comma we have:

e to the power of x equals 5 Replace u with e to the power of x in the equation u equals 5.
ln   e to the power of x equals ln   5 rightwards arrow x equals ln   5 Apply the natural logarithm to both sides of the equation e to the power of x equals 5 to solve for x. This is a possible solution.

For u equals short dash 1 comma we have:

e to the power of x equals short dash 1 Replace u with e to the power of x in the equation u equals short dash 1.
No solution The value of an exponential function cannot be negative, therefore there is no solution.

Thus, the only real solution to the equation is x equals ln     5.

summary
As you've learned up to this point, solving exponential equations can be performed using various methods. The one-to-one property of exponential functions can be applied when the bases are equal, or when the bases can be rewritten so that they are equal. However, when it is not possible to equate the bases, logarithms are used to solve the equation, a result of the one-to-one property of logarithms. You learned that you can solve equations involving one or two exponential functions by applying a logarithm to both sides—the common logarithm when the exponential function has base 10, the natural logarithm when it has base e, and either logarithm for all other bases. You also learned that when an equation has three terms, one-to-one properties will not help, so one strategy is to see if the equation is quadratic in form.

SOURCE: THIS WORK IS ADAPTED FROM PRECALCULUS BY JAY ABRAMSON. ACCESS FOR FREE AT OPENSTAX.ORG/BOOKS/PRECALCULUS/PAGES/1-INTRODUCTION-TO-FUNCTIONS

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