Exponential and Log Equations — AP® Precalculus Notes

When the unknown sits in an exponent or inside a logarithm, the usual algebra moves are not enough. The fix is the inverse pair from earlier topics: a logarithm undoes an exponential, and an exponential undoes a logarithm. This page solves both kinds of equations exactly, shows where extraneous solutions come from, handles equations that are quadratic in $e^x$ , and solves inequalities.

Exponential equations

An exponential equation has the unknown in an exponent. First isolate the power. Then either write both sides with the same base and set the exponents equal, or take a logarithm of both sides.

Worked example 1 (common base). Solve $4^{x + 1} = 8^x$ .

Both are powers of $2$ : $4^{x + 1}$ $= 2^{2(x + 1)}$ $= 2^{2x + 2}$ and $8^x = 2^{3x}$ .

Equal powers of $2$ have equal exponents: $2x + 2 = 3x$ , so $x = 2$ .

Worked example 2 (take a logarithm). Solve $5e^{3x} - 2 = 33$ exactly.

Isolate the power: $5e^{3x} = 35$ , so $e^{3x} = 7$ .

Take $\ln$ of both sides: $3x = \ln 7$ , so $x = \frac{\ln 7}{3}$ .

On a no-calculator question, leave the answer exact, as $\frac{\ln 7}{3}$ . Rounding it loses the point.

Logarithmic equations and extraneous solutions

A logarithmic equation has the unknown inside a logarithm. Use the properties of logarithms to get a single logarithm, then rewrite in exponential form.

Worked example 3. Solve $\log_3 x + \log_3(x - 8)$ $= 2$ .

Product property: $\log_3(x(x - 8)) = 2$ .

Exponential form: $x(x - 8) = 3^2 = 9$ , so $x^2 - 8x - 9 = 0$ and $(x - 9)(x + 1) = 0$ .

Check in the original. $x = 9$ : $\log_3 9 + \log_3 1$ $= 2 + 0$ $= 2$ . It works. $x = -1$ : $\log_3(-1)$ is undefined. It is an extraneous solution.

Answer: $x = 9$ .

Why do extra roots appear? The original equation needs $x > 0$ and $x - 8 > 0$ . The combined form $\log_3(x(x - 8))$ only needs the product to be positive, which also happens when both factors are negative. Combining logarithms quietly widened the domain.

Equations quadratic in $e^x$

Since $e^{2x} = (e^x)^2$ , an equation such as $e^{2x} - e^x - 12 = 0$ is a quadratic in disguise.

Worked example 4. Solve $e^{2x} - e^x - 12 = 0$ .

Let $u = e^x$ : $u^2 - u - 12 = 0$ , so $(u - 4)(u + 3) = 0$ and $u = 4$ or $u = -3$ .

$e^x = 4$ gives $x = \ln 4$ .

$e^x = -3$ has no solution, because $e^x > 0$ for every real $x$ . Say so in words when you reject it.

Inequalities

Applying an increasing function to both sides keeps the direction of an inequality; applying a decreasing one flips it. $\log_b$ and $b^x$ are increasing for $b > 1$ and decreasing for $0 < b < 1$ . Also keep every logarithm's input positive.

Inequality	Step	Solution
$2^x > 10$	apply $\log_2$ (increasing)	$x > \log_2 10$
$\left(\frac{1}{2}\right)^x < \frac{1}{8}$	$\frac{1}{8} = \left(\frac{1}{2}\right)^3$ , base below $1$ , flip	$x > 3$
$\log_2(x + 1) < 3$	$x + 1 < 8$ and $x + 1 > 0$	$-1 < x < 7$

Inverses of transformed functions

The same moves find inverses. For $f(x) = 5e^{2x} - 1$ : swap to $x = 5e^{2y} - 1$ , undo the outer steps in reverse order ( $e^{2y} = \frac{x + 1}{5}$ ), then take $\ln$ : $f^{-1}(x)$ $= \frac{1}{2}\ln \frac{x + 1}{5}$ . For $f(x) = \log_2(x + 1)$ : $x = \log_2(y + 1)$ gives $y + 1 = 2^x$ , so $f^{-1}(x) = 2^x - 1$ .