Solving Exponential Equations: Step by Step

Solving exponential equations means finding the exponent that makes the equation true. If both sides can be written as powers of the same base, set the exponents equal. If not, take the log of both sides, use the power rule to bring the variable down, and solve.

Before this lesson: You should know Properties of Logarithms, Logarithms and Exponent Rules.

What It Means to Solve for an Exponent

In an exponential equation, the variable is in an exponent, like $2^x = 10$ . Some are easy to solve by sight. You know $2^x = 8$ means $x = 3$ , because $2^3 = 8$ .

But $2^x = 10$ has no whole-number answer. Since $2^3 = 8$ and $2^4 = 16$ , the answer is between 3 and 4. Logarithms find the exact value: $x = \log_2 10 \approx 3.32$ . On a graph, it is where the curve $y = 2^x$ meets the line $y = 10$ .

Graph of the blue curve y = 2 to the x and the orange horizontal line y = 10, with x from −2 to 5 and y from 0 to 12. The curve passes through (0, 1), (2, 4) and (3, 8) and crosses the line at about (3.32, 10), which is marked with a point. — The solution of 2ˣ = 10 is the x-value where the two graphs meet, about 3.32.

Two Ways to Solve

Method 1: Same base. If you can write both sides as powers of the same base, the exponents must be equal:

\text{If } b^m = b^n, \text{ then } m = n.

Method 2: Take the log of both sides. Use this when the bases can’t match.

Get the power by itself on one side.
Take the log (or ln) of both sides.
Use the power rule to move the exponent to the front: $\log b^x = x \log b$ .
Solve for $x$ . Divide by the log, then do any other steps.
Use a calculator only at the end, and round as asked.

Calculator note: For $\frac{\log 40}{\log 5}$ , press LOG, 40, ), ÷, LOG, 5, ), then ENTER.

Worked Examples

Example 1: Same base on both sides

Solve $3^{x + 1} = 81$ .

81 = 3^4

Write 81 as a power of 3.

3^{x + 1} = 3^4

Rewrite the equation.

x + 1 = 4

Set the exponents equal.

x = 3

Subtract 1 from both sides.

Check: $3^{3 + 1} = 3^4 = 81$

Answer: $x = 3$

Example 2: Rewrite both bases

Solve $4^x = 8^{x - 1}$ .

Write each base as a power of 2: $4 = 2^2$ and $8 = 2^3$

(2^2)^x = (2^3)^{x - 1}

Rewrite the equation.

2^{2x} = 2^{3x - 3}

Multiply the exponents (see power of a power rule)

2x = 3x - 3

Set the exponents equal.

-x = -3

Subtract $3x$ from both sides.

x = 3

Divide both sides by −1.

Check: $4^3 = 64$ and $8^{3 - 1} = 8^2 = 64$

Answer: $x = 3$

Example 3: Take the log of both sides

Solve $5^x = 40$ . Round to the nearest hundredth.

\log 5^x = \log 40

40 is not a power of 5, so take the log of both sides.

x \log 5 = \log 40

Use the power rule (see properties of logarithms)

x = \frac{\log 40}{\log 5}

Divide both sides by $\log 5$ .

Evaluate each log: $\log 40 \approx 1.60206$ and $\log 5 \approx 0.69897$

\frac{1.60206}{0.69897} \approx 2.292

Divide.

x \approx 2.29

Round to the nearest hundredth.

Check: $5^{2.29} \approx 39.9$ , which is close to 40.

Answer: $x \approx 2.29$

Example 4: Isolate the power first

Solve $2 \cdot 7^{x - 3} + 5 = 41$ . Round to the nearest hundredth.

2 \cdot 7^{x - 3} = 36

Subtract 5 from both sides.

7^{x - 3} = 18

Divide both sides by 2.

\log 7^{x - 3} = \log 18

Take the log of both sides.

(x - 3) \log 7 = \log 18

Use the power rule.

x - 3 = \frac{\log 18}{\log 7}

Divide both sides by $\log 7$ .

Evaluate each log: $\log 18 \approx 1.25527$ and $\log 7 \approx 0.84510$

\frac{1.25527}{0.84510} \approx 1.4854

Divide.

x \approx 1.4854 + 3 = 4.4854

Add 3 to both sides.

x \approx 4.49

Round to the nearest hundredth.

Answer: $x \approx 4.49$

Example 5: Word problem with population growth

A town has 12,000 people, and its population grows 3% each year. Its population after $t$ years is $P = 12{,}000(1.03)^t$ . After how many years will the town reach 18,000 people?

12{,}000(1.03)^t = 18{,}000

Set $P$ equal to 18,000.

1.03^t = 1.5

Divide both sides by 12,000.

\log 1.03^t = \log 1.5

Take the log of both sides.

t \log 1.03 = \log 1.5

Use the power rule.

t = \frac{\log 1.5}{\log 1.03}

Divide both sides by $\log 1.03$ .

Evaluate each log: $\log 1.5 \approx 0.176091$ and $\log 1.03 \approx 0.012837$

\frac{0.176091}{0.012837} \approx 13.72

Divide.

Answer: about 13.7 years, so the town passes 18,000 people during year 14. The same steps work for compound interest problems.

Common Mistakes

Mistake	Example of the error	Fix
Taking the log before isolating the power	$\log (2^x + 5) = x \log 2 + \log 5$	Subtract 5 first. There is no rule for the log of a sum.
Combining the coefficient with the base	$3 \cdot 2^x = 48$ becomes $6^x = 48$	Divide by 3 first: $2^x = 16$ .
Turning a quotient of logs into one log	$\frac{\log 40}{\log 5} = \log 8$	The quotient rule is about $\log \frac{M}{N}$ . Divide the two decimal values instead.
Setting exponents equal when the bases differ	$2^x = 3^{x - 1}$ , so $x = x - 1$	Only equal bases let you match exponents. Take logs instead.
Rounding too early	Using $\log 7 \approx 0.8$	Keep at least four decimal places until the last step.

Related Methods

Use ln when the base is e

If the base is the number e, take the natural log of both sides, because $\ln e^x = x$ . For $e^{2x} = 30$ , you get $2x = \ln 30$ , so $x = \frac{\ln 30}{2} \approx 1.70$ .

Write the answer as a log

The exact answer to $5^x = 40$ is $x = \log_5 40$ . The change of base formula turns it into $\frac{\log 40}{\log 5}$ , the same value as Method 2.

Solve by graphing

Graph each side as its own function, $y = 5^x$ and $y = 40$ , and find where they meet. This gives a good estimate and a check on your algebra.

Practice Problems

Solve $2^x = 32$ .
Solve $5^{2x - 1} = 125$ .
Solve $9^x = 27$ .
Solve $(\frac{1}{2})^x = 8$ .
Solve $3^x = 20$ . Round to the nearest hundredth.
Solve $6^{x + 2} = 50$ . Round to the nearest hundredth.
Solve $4 \cdot 2^x = 100$ . Round to the nearest hundredth.
Solve $e^x = 15$ . Round to the nearest hundredth.
Solve $3e^{2x} - 4 = 20$ . Round to the nearest hundredth.
A lab starts with 500 bacteria, and the number doubles every hour: $N = 500 \cdot 2^t$ . After how many hours will there be 20,000 bacteria?

Show answers

32 = 2^5

Write 32 as a power of 2.

2^x = 2^5

Rewrite.

x = 5

Set the exponents equal.

Answer: $x = 5$

125 = 5^3

Write 125 as a power of 5.

2x - 1 = 3

Set the exponents equal.

2x = 4

Add 1 to both sides.

x = 2

Divide both sides by 2.

Answer: $x = 2$

Write each base as a power of 3: $9 = 3^2$ and $27 = 3^3$

(3^2)^x = 3^3

Rewrite.

3^{2x} = 3^3

Multiply the exponents.

2x = 3

Set the exponents equal.

x = \frac{3}{2}

Divide both sides by 2.

Answer: $x = \frac{3}{2}$

Write each side as a power of 2: $\frac{1}{2} = 2^{-1}$ and $8 = 2^3$

(2^{-1})^x = 2^3

Rewrite.

2^{-x} = 2^3

Multiply the exponents.

-x = 3

Set the exponents equal.

x = -3

Divide both sides by −1.

Answer: $x = -3$

\log 3^x = \log 20

Take the log of both sides.

x \log 3 = \log 20

Use the power rule.

x = \frac{\log 20}{\log 3}

Divide by $\log 3$ .

\frac{1.30103}{0.47712} \approx 2.727

Evaluate.

Answer: $x \approx 2.73$

\log 6^{x + 2} = \log 50

Take the log of both sides.

(x + 2) \log 6 = \log 50

Use the power rule.

x + 2 = \frac{\log 50}{\log 6}

Divide by $\log 6$ .

\frac{1.69897}{0.77815} \approx 2.1833

Evaluate.

x \approx 2.1833 - 2 = 0.1833

Subtract 2 from both sides.

Answer: $x \approx 0.18$

2^x = 25

Divide both sides by 4.

\log 2^x = \log 25

Take the log of both sides.

x \log 2 = \log 25

Use the power rule.

x = \frac{\log 25}{\log 2}

Divide by $\log 2$ .

\frac{1.39794}{0.30103} \approx 4.644

Evaluate.

Answer: $x \approx 4.64$

\ln e^x = \ln 15

Take the natural log of both sides.

x = \ln 15

Use $\ln e^x = x$ .

\ln 15 \approx 2.708

Evaluate.

Answer: $x \approx 2.71$

3e^{2x} = 24

Add 4 to both sides.

e^{2x} = 8

Divide both sides by 3.

2x = \ln 8

Take the natural log of both sides.

x = \frac{\ln 8}{2}

Divide both sides by 2.

\frac{2.07944}{2} = 1.03972

Evaluate.

Answer: $x \approx 1.04$

10.

500 \cdot 2^t = 20{,}000

Set $N$ equal to 20,000.

2^t = 40

Divide both sides by 500.

t \log 2 = \log 40

Take the log of both sides.

t = \frac{\log 40}{\log 2}

Divide by $\log 2$ .

\frac{1.60206}{0.30103} \approx 5.32

Evaluate.

Answer: after about 5.32 hours

FAQ

How do you solve an exponential equation without logs?

Write both sides as powers of the same base, then set the exponents equal. For $2^{x + 3} = 2^7$ , you get $x + 3 = 7$ , so $x = 4$ .

Should I use log or ln?

Either one works for any base, and both give the same answer. Use ln when the base is e, because $\ln e^x = x$ saves a step.

Why do you take the log of both sides?

Doing the same thing to both sides keeps the equation true. The power rule then moves the variable out of the exponent, so you can solve for it with division.

Can an exponential equation have no solution?

Yes. A positive base raised to any power is always positive, so an equation like $2^x = -4$ has no real solution.

Summary

If both sides are powers of the same base, set the exponents equal.
Otherwise, isolate the power, then take the log of both sides.
The power rule turns $\log b^x$ into $x \log b$ , which frees the variable.
Use ln when the base is e.
Round only at the end, and check your answer in the original equation.

Previous lesson: Properties of Logarithms

Next lesson: Solving Logarithmic Equations

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