How do I solve an exponential equation without a calculator?

If both sides can be written as powers of the same base, set the exponents equal. Otherwise, isolate the exponential part and take the natural log of both sides.

What is an extraneous solution of a log equation?

A value that comes out of the algebra but makes the argument of one of the original logs zero or negative. It isn't a solution, so you reject it.

How do I change a yearly growth factor to a monthly one?

Raise the yearly factor to the power 1/12. A yearly factor of 1.08 becomes a monthly factor of about 1.0064.

Does Desmos use base 10 or base e for log?

In Desmos, log means base 10 and ln means base e. For another base, type it as a subscript, such as log_2(x).

Solving exponential and logarithmic equations for AP Precalculus

Updated November 10, 2026

Much of AP Precalculus Unit 2 comes down to a handful of rules for exponents and logarithms, used in two ways: to rewrite an expression in an equivalent form, and to solve an equation exactly. Both show up in no-calculator multiple choice and in the fourth free-response question, Symbolic Manipulations.

This guide lists the rules you need, shows how to rewrite a growth model for a different time unit, and works through exponential and log equations by hand, including how to spot and reject extraneous solutions.

The rules you need

For a positive base $b$ , the exponent rules are:

Product: $b^{m} \cdot b^{n} = b^{m + n}$
Quotient: $\frac{b^{m}}{b^{n}} = b^{m - n}$
Power: $(b^{m})^{n} = b^{mn}$
Negative and fractional exponents: $b^{-n} = \frac{1}{b^{n}}$ and $b^{\frac{1}{n}} = \sqrt[n]{b}$

A logarithm is an exponent. $\log_b(x) = c$ means exactly $b^{c} = x$ , for $b > 0$ , $b \ne 1$ , and $x > 0$ . The log rules are the exponent rules read backward:

Product: $\log_b(MN) = \log_b M + \log_b N$
Quotient: $\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N$
Power: $\log_b(M^{k}) = k \log_b M$
Change of base: $\log_b x = \frac{\ln x}{\ln b}$
Inverses: $\log_b(b^{x}) = x$ and $b^{\log_b x} = x$

$\log$ with no base means base $10$ , and $\ln$ means base $e$ . The same is true in Desmos, which also accepts other bases typed like $\log_{2}(x)$ .

Rewriting a model in an equivalent form

The power rule lets you change the time unit of an exponential model without changing the model. If a quantity grows by a factor of $1.08$ per year, then over $t$ years it grows by $1.08^{t}$ . To write it per month, use $t = \frac{m}{12}$ : $1.08^{\frac{m}{12}} = (1.08^{\frac{1}{12}})^{m} \approx 1.0064^{m}$ . Per decade, it's $(1.08^{10})^{d} \approx 2.159^{d}$ .

A half-life: $80\left(\frac{1}{2}\right)^{\frac{t}{5}} = 80\left(\left(\frac{1}{2}\right)^{\frac{1}{5}}\right)^{t} \approx 80(0.871)^{t}$ . Halving every $5$ years is the same as losing about $12.9\%$ per year.
Base $e$ to another base: $e^{0.3t} = (e^{0.3})^{t} \approx 1.350^{t}$ .
A shift is a stretch: $2^{x + 3} = 2^{3} \cdot 2^{x} = 8 \cdot 2^{x}$ . A horizontal shift of an exponential graph is the same as a vertical stretch.

From decades to years

A town's population is modeled by

P(t) = 4200(1.15)^{t}

, where

t

is the number of decades since 2000. (a) Rewrite the model as

P = 4200 \cdot b^{y}

, where

y

is the number of years since 2000. (b) By about what percent does the population grow each year? (c) What does the model give for the population in 2030?

(a) One decade is $10$ years, so $t = \frac{y}{10}$ . Then $1.15^{\frac{y}{10}} = \left(1.15^{\frac{1}{10}}\right)^{y}$ , and $1.15^{\frac{1}{10}} \approx 1.0141$ . So $P \approx 4200(1.0141)^{y}$ .

(b) A factor of about $1.0141$ per year means growth of about $1.41\%$ per year. It's not $1.5\%$ : dividing the $15\%$ by $10$ ignores the compounding.

Answer: (a) $P = 4200\left(1.15^{\frac{1}{10}}\right)^{y} \approx 4200(1.0141)^{y}$ . (b) About $1.41\%$ per year. (c) About $6388$ people.

Solving exponential equations

Exponential functions are one-to-one, so if $b^{M} = b^{N}$ then $M = N$ . That gives two methods.

Match the bases when both sides are powers of the same number, then set the exponents equal.
Isolate, then take a log when they aren't. Get the exponential part alone on one side first, then take $\ln$ (or the log with a matching base) of both sides.

Matching bases

Solve

8^{x + 1} = 16^{x - 2}

exactly.

Both bases are powers of $2$ : $8 = 2^{3}$ and $16 = 2^{4}$ . So the equation is $2^{3(x + 1)} = 2^{4(x - 2)}$ , or $2^{3x + 3} = 2^{4x - 8}$ .

Set the exponents equal: $3x + 3 = 4x - 8$ , so $x = 11$ . Check: $8^{12} = 2^{36}$ and $16^{9} = 2^{36}$ .

Answer: $x = 11$

Isolate before you take the log

Solve

3e^{2x} - 5 = 16

exactly.

Add $5$ and divide by $3$ : $e^{2x} = 7$ . Now take $\ln$ of both sides: $2x = \ln 7$ , so $x = \frac{\ln 7}{2}$ .

Taking $\ln$ first would give $\ln(3e^{2x} - 5)$ , which doesn't simplify. For reference, $x \approx 0.973$ , but on a no-calculator question the exact form is the answer.

Answer: $x = \frac{\ln 7}{2}$

Two different bases

Solve

4^{x} = 3^{x + 2}

exactly.

The bases don't match, so take $\ln$ of both sides and use the power rule: $x \ln 4 = (x + 2) \ln 3 = x \ln 3 + 2 \ln 3$ .

Collect the $x$ terms and factor: $x(\ln 4 - \ln 3) = 2 \ln 3$ . Divide: $x = \frac{2 \ln 3}{\ln 4 - \ln 3}$ , which can also be written $\frac{\ln 9}{\ln(4/3)}$ . That's about $7.638$ .

Answer: $x = \frac{2 \ln 3}{\ln 4 - \ln 3}$

A quadratic in e to the x

Solve

e^{2x} - 2e^{x} - 15 = 0

exactly.

Since $e^{2x} = (e^{x})^{2}$ , let $u = e^{x}$ . The equation becomes $u^{2} - 2u - 15 = 0$ , which factors as $(u - 5)(u + 3) = 0$ .

So $e^{x} = 5$ or $e^{x} = -3$ . An exponential is always positive, so $e^{x} = -3$ has no solution. From $e^{x} = 5$ , $x = \ln 5$ .

Answer: $x = \ln 5$ (the case $e^{x} = -3$ has no solution)

Solving log equations and checking for extraneous solutions

1Combine the logs on each side into a single log, using the product, quotient, and power rules.
2Rewrite in exponential form ( $\log_b A = c$ becomes $A = b^{c}$ ), or, if both sides are single logs with the same base, set the arguments equal.
3Solve the equation that's left.
4Check every answer in the original equation. Each argument of each log must be positive. Reject any value that fails, and say why.

The check matters because combining logs can widen the domain. $\log_2(x) + \log_2(x - 6)$ needs $x > 0$ and $x > 6$ , but $\log_2(x(x - 6))$ is also defined when $x < 0$ . Algebra that's correct can still hand you a value the original equation never allowed.

One answer is extraneous

Solve

\log_2(x) + \log_2(x - 6) = 4

Combine: $\log_2(x(x - 6)) = 4$ . Rewrite: $x(x - 6) = 2^{4} = 16$ , so $x^{2} - 6x - 16 = 0$ , which factors as $(x - 8)(x + 2) = 0$ .

Check $x = 8$ : $\log_2 8 + \log_2 2 = 3 + 1 = 4$ . It works. Check $x = -2$ : $\log_2(-2)$ is undefined, so $x = -2$ is extraneous.

Answer: $x = 8$

A quotient of logs

Solve

\ln(x + 4) - \ln(x - 2) = \ln 3

Combine the left side: $\ln\left(\frac{x + 4}{x - 2}\right) = \ln 3$ . Both sides are single natural logs, so the arguments are equal: $\frac{x + 4}{x - 2} = 3$ .

Multiply by $x - 2$ : $x + 4 = 3x - 6$ , so $x = 5$ . Check: $x + 4 = 9$ and $x - 2 = 3$ are both positive, and $\ln 9 - \ln 3 = \ln 3$ .

Answer: $x = 5$

Isolate the log first

Solve

3 + 2\log_5(x - 1) = 7

Subtract $3$ and divide by $2$ : $\log_5(x - 1) = 2$ . Rewrite: $x - 1 = 5^{2} = 25$ , so $x = 26$ .

Check: $x - 1 = 25 > 0$ , and $3 + 2\log_5 25 = 3 + 2(2) = 7$ .

Answer: $x = 26$

A quick check

Which values solve it

Solve

\log_3(x + 1) + \log_3(x - 1) = 1

A. $x = 2$ and $x = -2$
B. $x = 2$ only
C. $x = \frac{3}{2}$
D. $x = -2$ only

Combine: $\log_3((x + 1)(x - 1)) = 1$ , so $x^{2} - 1 = 3$ and $x = \pm 2$ . The value $x = -2$ makes both arguments negative, so it's extraneous. Check $x = 2$ : $\log_3 3 + \log_3 1 = 1 + 0 = 1$ . Choice C comes from adding the arguments instead of multiplying them.

Answer: B

When an equation mixes an exponential or log with another kind of function, like $2^{x} = 3x$ , no rule isolates $x$ . On a calculator part, graph both sides in Desmos and click the intersections; see Desmos on the AP Precalculus exam. To see these rules at work on real data, read the regression guide.

Keep going

Practice on exam-style questions

Intersect Prep has 112 multiple-choice questions for Unit 2, with a full solution for each one. Create a free account to try a full lesson with the calculator. One payment of $29 unlocks the course for 3 months, or $39 for 6 months.

Try a free lesson See the course

Questions

How do I solve an exponential equation without a calculator?: If both sides can be written as powers of the same base, set the exponents equal. Otherwise, isolate the exponential part and take the natural log of both sides.
What is an extraneous solution of a log equation?: A value that comes out of the algebra but makes the argument of one of the original logs zero or negative. It isn't a solution, so you reject it.
How do I change a yearly growth factor to a monthly one?: Raise the yearly factor to the power 1/12. A yearly factor of 1.08 becomes a monthly factor of about 1.0064.
Does Desmos use base 10 or base e for log?: In Desmos, log means base 10 and ln means base e. For another base, type it as a subscript, such as log_2(x).