What is the first thing to try when solving an exponential equation?

Check whether both sides can be written with the same base. If so, set the exponents equal and solve, with no logarithm needed. For example, $2^x = 16$ becomes $2^x = 2^4$, so $x = 4$. Only when the bases cannot be matched do you take a logarithm of both sides to bring the variable exponent down.

How do you free a variable that is stuck in an exponent?

Take a logarithm of both sides and apply the power property, which moves the exponent down as a coefficient. For $5^x = 30$, take the natural log: $\ln 5^x = \ln 30$, so $x\ln 5 = \ln 30$ and $x = \frac{\ln 30}{\ln 5}$. Either the natural log or the common log works, since you are just inverting the exponential.

How do you solve a logarithmic equation?

Condense the equation into a single logarithm using the product, quotient and power properties, then rewrite $\log_b(\text{argument}) = k$ in exponential form as argument $= b^k$. Solve the resulting algebraic equation, then check every candidate, because the argument of a logarithm must be positive.

Why do logarithmic equations need an extraneous-solution check?

A logarithm is defined only for positive arguments. The algebra of condensing and exponentiating can produce candidate solutions that make an argument zero or negative, which are not in the domain. Each candidate must be substituted back to confirm every logarithm's argument is positive; any that fail are extraneous and are rejected.

Why does AP Precalculus expect exact answers on the no-calculator section?

The no-calculator parts test whether you can manipulate exponentials and logarithms exactly, not just read a number off a screen. You are expected to show the steps - matching bases, applying the power property, rewriting in exponential form - that justify the solution. The exam awards the method, so a bare answer can lose points and correct work with a slip can still earn partial credit.

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AP Precalculus: a complete guide to solving exponential and logarithmic equations on the exam

A deep-dive AP Precalculus guide to solving exponential and logarithmic equations. Covers matching bases, taking logs to free an exponent, condensing and exponentiating to free a logarithm, the change-of-base and log properties, checking for extraneous solutions, and the no-calculator exam technique the College Board rewards.

Generated by Claude Opus 4.818 min read2.3-2.13Updated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What solving these equations actually demands
Match the inverse operation to the equation
Exponential equations: try matching bases first
Exponential equations: take a log when bases will not match
Logarithmic equations: condense, then exponentiate
Always check for extraneous solutions
Inequalities
How these equations are examined
Check your knowledge

What solving these equations actually demands

Solving exponential and logarithmic equations is a core skill of AP Precalculus Unit 2, and both sections lean on it. The College Board wants you to find an exact value when you can, choose the right inverse operation, apply the logarithm properties correctly, and check for extraneous solutions. This guide ties together the matching dot-point pages, each with its own practice: exponential functions, exponential function manipulation, logarithmic expressions, logarithmic function manipulation, exponential and logarithmic equations and inequalities, and inverses of exponential functions.

Match the inverse operation to the equation

Every exponential or logarithmic equation hides the variable somewhere, and the fix is the inverse operation that frees it.

Variable in an exponent (like $3^x = 20$ ): take a logarithm of both sides.
Variable inside a logarithm (like $\log_2(x) = 5$ ): exponentiate both sides (rewrite in exponential form).

Because the exponential and the logarithm are inverses, each one undoes the other. Recognizing which case you are in tells you which move to make, so you never have to guess.

Exponential equations: try matching bases first

Before reaching for a logarithm, check whether both sides can be written with the same base. If they can, set the exponents equal.

8^{x} = 32 \;\Rightarrow\; (2^3)^x = 2^5 \;\Rightarrow\; 2^{3x} = 2^{5} \;\Rightarrow\; 3x = 5 \;\Rightarrow\; x = \tfrac{5}{3}.

Matching bases gives an exact answer with no logarithm, which is ideal on the no-calculator section.

Exponential equations: take a log when bases will not match

When the bases cannot be matched, take a logarithm of both sides and use the power property $\log(b^k) = k\log b$ to bring the exponent down.

5^{x} = 30 \;\Rightarrow\; \ln 5^{x} = \ln 30 \;\Rightarrow\; x\ln 5 = \ln 30 \;\Rightarrow\; x = \frac{\ln 30}{\ln 5}.

Either $\ln$ or $\log$ works; the change-of-base idea means $\frac{\ln 30}{\ln 5} = \log_5 30$ . On the calculator section, evaluate; on the no-calculator section, leave the exact logarithmic form.

Logarithmic equations: condense, then exponentiate

For a logarithmic equation, first condense to a single logarithm with the product, quotient and power properties, then rewrite in exponential form to remove the logarithm.

\log_2(x) + \log_2(x - 2) = 3 \;\Rightarrow\; \log_2\big(x(x-2)\big) = 3 \;\Rightarrow\; x(x-2) = 2^3 = 8.

This leaves an ordinary algebraic equation, $x^2 - 2x - 8 = 0$ , which factors to $(x-4)(x+2) = 0$ .

Always check for extraneous solutions

A logarithm requires a positive argument, so every candidate must be checked. In the example above, $x = 4$ and $x = -2$ both solve the quadratic, but $x = -2$ makes $\log_2(-2)$ undefined. So $x = -2$ is extraneous and the only valid solution is $x = 4$ .

A full logarithmic equation with a check

Solve $\log_5(2x + 3) - \log_5(x - 1) = 1$ with complete justification.

step 1 Condense the left side

Use the quotient property: $\log_5\dfrac{2x + 3}{x - 1} = 1$ .

step 2 Rewrite in exponential form

$\log_5(\text{argument}) = 1$ means $\text{argument} = 5^1 = 5$ , so $\dfrac{2x + 3}{x - 1} = 5$ .

step 3 Solve the algebraic equation

Multiply by $x - 1$ : $2x + 3 = 5(x - 1) = 5x - 5$ . Then $3 + 5 = 5x - 2x$ , so $8 = 3x$ and $x = \dfrac{8}{3}$ .

step 4 Check the domain

The arguments are $2x + 3 = \frac{16}{3} + 3 > 0$ and $x - 1 = \frac{8}{3} - 1 = \frac{5}{3} > 0$ , both positive. So $x = \dfrac{8}{3}$ is valid.

Inequalities

An exponential or logarithmic inequality is solved like the matching equation, with one extra concern: the direction. For a base greater than $1$ , the exponential and logarithmic functions are increasing, so taking a log or exponentiating preserves the inequality direction. You then intersect the algebraic solution with the domain: every logarithm argument must stay positive, which can trim the solution set. For instance, solving $\log(x) < 2$ gives $x < 10^2 = 100$ , but combined with the domain $x > 0$ , the solution is $0 < x < 100$ .

How these equations are examined

Matching bases. Rewrite both sides to a common base and set exponents equal, for exact no-calculator answers.
Taking logs. Free a variable exponent with the power property; leave exact logarithmic form on the no-calculator part.
Condensing and exponentiating. Combine logs into one, rewrite in exponential form, and solve the resulting equation.
Extraneous checks. Reject any candidate that makes a logarithm argument non-positive.

Check your knowledge

A mix of exponential and logarithmic equations. Work them under no-calculator conditions where possible, then check against the solutions.

Solve $2^{x} = 64$ . (1 mark)
Solve $9^{x} = 27$ by matching bases. (2 marks)
Solve $4^{x} = 10$ , leaving an exact logarithmic answer. (2 marks)
Solve $\log_3(x) = 4$ . (1 mark)
Solve $\log(x) + \log(x - 3) = 1$ , checking for extraneous solutions. (3 marks)
Solve $\ln(x) = 0$ . (1 mark)
Solve $5^{x+1} = 25^{x}$ by matching bases. (2 marks)
Solve $\log_2(x + 6) - \log_2(x) = 2$ , checking the domain. (3 marks)

Solutions

Step 1: Match the base for Q1

Both sides can be written as powers of $2$ , so rewrite $64 = 2^6$ and set exponents equal. Since $2^x = 2^6$ , we have $x = 6$ .

Step 2: Match the base for Q2

Rewrite both sides as powers of $3$ : $9 = 3^2$ and $27 = 3^3$ , so $(3^2)^x = 3^3$ gives $3^{2x} = 3^3$ . Setting exponents equal: $2x = 3$ , so $x = \dfrac{3}{2}$ .

Step 3: Take the natural log for Q3

The bases $4$ and $10$ cannot be matched, so take $\ln$ of both sides and apply the power property to bring the exponent down. This gives $x \ln 4 = \ln 10$ , so $x = \dfrac{\ln 10}{\ln 4} = \log_4 10$ .

Step 4: Rewrite in exponential form for Q4

The equation $\log_3(x) = 4$ means $x = 3^4 = 81$ . Converting from logarithmic to exponential form undoes the logarithm directly.

Step 5: Condense and check the domain for Q5

Use the product property to condense: $\log\big(x(x-3)\big) = 1$ , so $x(x-3) = 10^1 = 10$ . This gives $x^2 - 3x - 10 = 0$ , which factors as $(x-5)(x+2) = 0$ , yielding candidates $x = 5$ and $x = -2$ . Both logarithm arguments require $x > 3$ , so $x = -2$ is extraneous and the solution is $x = 5$ .

Step 6: Rewrite in exponential form for Q6

The natural logarithm $\ln(x) = 0$ means $x = e^0 = 1$ , since any base raised to the power zero equals $1$ .

Step 7: Match the base for Q7

Rewrite $25 = 5^2$ so that both sides share base $5$ : $5^{x+1} = 5^{2x}$ . Setting the exponents equal gives $x + 1 = 2x$ , so $x = 1$ .

Step 8: Condense and verify the domain for Q8

Use the quotient property: $\log_2\!\dfrac{x+6}{x} = 2$ , so $\dfrac{x+6}{x} = 2^2 = 4$ . Solving: $x + 6 = 4x$ , so $6 = 3x$ and $x = 2$ . Check the domain: both $x + 6 = 8 > 0$ and $x = 2 > 0$ , so $x = 2$ is valid.

Sources & how we know this

AP Precalculus Course and Exam Description — College Board (2023)