Exponential and logarithmic equations and inequalities — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: One-to-one properties for exponentials and logarithms, solving exponential equations via common base and logarithm methods, solving logarithmic equations, identifying extraneous solutions, solving exponential and logarithmic inequalities.

You should already know: Properties of exponents and logarithms; how to solve linear and quadratic equations; domain restrictions for logarithmic functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Exponential and logarithmic equations and inequalities?

Exponential equations are equations where the unknown variable appears in the exponent of an exponential expression; logarithmic equations are equations where the unknown appears in the argument or base of a logarithmic expression. Corresponding inequalities have the same structure but use inequality symbols instead of equals signs. This topic is part of Unit 2: Exponential and Logarithmic Functions, which accounts for 28-35% of the total AP Precalculus exam score, with this subtopic making up roughly 8-10% of the exam. Questions on this topic appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often embedded in real-world modeling problems. Solving these equations relies on the inverse relationship between exponential and logarithmic functions, and requires careful attention to domain restrictions that can lead to invalid extraneous solutions. Inequalities require additional attention to how the base of the function impacts inequality direction.

2. Solving Exponential Equations

An exponential equation has the general form $a^{f(x)}=b^{g(x)}$, where $a,b>0$ and $a,b\ne 1$. There are two core methods for solving these equations, depending on whether the bases can be rewritten as a common base. The first method uses the one-to-one property of exponential functions: for $a>0,a\ne 1$, if $a^m=a^n$, then $m=n$. If both sides of the equation can be rewritten with the same base, we can directly drop the base and solve the resulting equation for $x$. If the bases cannot be matched, we take the natural logarithm (or common logarithm) of both sides, then apply the power rule for logarithms $\ln a^b=b\ln a$ to bring the exponent down to a linear term, then solve for $x$. For the general case $a^{kx+c}=d$, the solution can be derived as: $$(kx+c)\ln a=\ln d⟹x=\frac{\ln d-c\ln a}{k\ln a}$$ This gives an exact solution, which can be approximated with a calculator for a decimal value.

Worked Example

Solve $2^{4x+3}=5^{x-2}$ for $x$, giving both exact and approximate values to 3 decimal places.

1. The bases 2 and 5 are distinct and cannot be rewritten as a common base, so we use the logarithm method.
2. Take the natural logarithm of both sides: $\ln \left(2^{4x+3}\right)=\ln \left(5^{x-2}\right)$.
3. Apply the power rule to bring down exponents: $(4x+3)\ln 2=(x-2)\ln 5$.
4. Expand and collect like terms: $4x\ln 2+3\ln 2=x\ln 5-2\ln 5⟹x(4\ln 2-\ln 5)=-2\ln 5-3\ln 2$.
5. Solve for $x$: $$x=\frac{-(2\ln 5+3\ln 2)}{4\ln 2-\ln 5}\approx \frac{-(3.2189+2.0794)}{2.7726-1.6094}\approx -4.553$$ Exact form: $x=\frac{-(2\ln 5+3\ln 2)}{4\ln 2-\ln 5}$, approximate: $x\approx -4.553$.

Exam tip: Always confirm what form the question asks for (exact vs approximate); AP Precalc almost always requires 3 decimal places for approximate answers, so double-check your rounding.

3. Solving Logarithmic Equations

A logarithmic equation has the unknown variable in the argument or base of a logarithm. Like exponential equations, we use the one-to-one property and inverse relationship between exponentials and logarithms to solve these. If you have two logarithms of the same base on opposite sides of the equals sign, the one-to-one property says if ${\log }_{b}A={\log }_{b}C$, then $A=C$ (for $b>0,b\ne 1$). If you have a single logarithm equal to a constant, rewrite the equation in exponential form: ${\log }_{b}A=k⟺b^k=A$. The most critical step that many students miss is checking for extraneous solutions: the argument of every logarithm in the original equation must be strictly positive, so any solution that makes an argument non-positive must be discarded. This is required even if the question does not explicitly ask you to check solutions.

Worked Example

Solve ${\log }_2(x+2)+{\log }_2(x-1)=2$.

1. First, note the domain restrictions from the original equation: $x+2>0⟹x>-2$, and $x-1>0⟹x>1$, so the domain is $x>1$.
2. Use the product rule for logarithms to combine the two terms: ${\log }_2\left[(x+2)(x-1)\right]=2$.
3. Rewrite in exponential form using the definition of logarithms: $(x+2)(x-1)=2^2=4$.
4. Expand and rearrange into a standard quadratic equation: $x^2+x-2=4⟹x^2+x-6=0$.
5. Factor and solve: $(x+3)(x-2)=0⟹x=-3$ or $x=2$.
6. Check against the domain: $x=-3<1$, so it is extraneous and discarded. $x=2>1$ satisfies the domain, so it is the only valid solution.

Exam tip: Always write domain restrictions from the original equation, not just the combined logarithm; it is possible for the combined argument to be positive even if an original argument is negative, leading to an invalid solution.

4. Solving Exponential and Logarithmic Inequalities

Exponential and logarithmic inequalities follow the same initial steps as equations: find the domain, rewrite to get a common base, then drop the base or logarithm to get a polynomial/rational inequality. The key difference is that inequality direction changes based on the monotonicity (increasing/decreasing behavior) of the function:

• For base $b>1$: $b^x$ and ${\log }_{b}x$ are strictly increasing, so inequality direction is preserved when dropping the base/logarithm.
• For $0<b<1$: $b^x$ and ${\log }_{b}x$ are strictly decreasing, so inequality direction reverses when dropping the base/logarithm. After finding the solution to the simplified inequality, you intersect this solution with the domain of the original inequality to get the final solution set.

Worked Example

Solve ${\log }_{0.5}(2x-3)<{\log }_{0.5}(x+1)$.

1. Find the domain: $2x-3>0⟹x>1.5$, and $x+1>0⟹x>-1$, so the domain is $x>1.5$.
2. Both sides are logarithms with the same base $b=0.5$, which is between 0 and 1, so ${\log }_{0.5}x$ is strictly decreasing. This means we reverse the inequality sign when we drop the logarithm.
3. Rewrite the inequality: $2x-3>x+1$.
4. Solve the simplified inequality: $x>4$.
5. Intersect with the domain $x>1.5$: the final solution is $x>4$, or $(4,\infty )$ in interval notation.

Exam tip: If the question does not specify a form for the solution, use interval notation; it is universally accepted on the AP Precalc exam and less prone to notation errors.

5. Common Pitfalls (and how to avoid them)

• Wrong move: When solving ${\log }_{b}A={\log }_{b}C$, immediately conclude $A=C$ and keep all solutions to $A=C$ without checking domain. Why: Students remember the one-to-one property but forget that only positive arguments are valid, so extraneous solutions are often left in. Correct move: Always write down the domain of the original equation before starting to solve, and discard any solution that does not satisfy the domain restriction.
• Wrong move: When solving $2^x<2^{x-5}$, automatically reverse the inequality sign because it is an inequality. Why: Students confuse the base rule, reversing the sign regardless of whether the base is greater than 1. Correct move: Before dropping the base, explicitly check the base: if $b>1$, keep inequality direction; if $0<b<1$, reverse it.
• Wrong move: When solving $5^{2x}=3^x$, expand to $2x\ln 5=x\ln 3$, then divide both sides by $x$ to get $x=\frac{\ln 3}{2\ln 5}$, missing the solution $x=0$. Why: Students forget dividing by a variable expression assumes it is non-zero, losing the root at zero. Correct move: Factor out the common variable term instead of dividing: $x(2\ln 5-\ln 3)=0$, which captures both solutions.
• Wrong move: When rewriting $\ln (2x+1{)}^2=6$, apply the power rule to get $2\ln (2x+1)=6$, leading to only one solution. Why: The power rule $\ln a^b=b\ln a$ only holds when $a>0$, and squaring makes the argument positive even if $2x+1$ is negative. Correct move: Rewrite in exponential form first: $(2x+1{)}^2=e^6$, so $2x+1=\pm e^3$, giving both valid solutions.
• Wrong move: When solving $\ln x+\ln (x-2)=3$, combine to $\ln [x(x-2)]=3$, solve, and keep any solution that makes $x(x-2)>0$. Why: The combined argument can be positive for $x<0$, which still makes $\ln (x-2)$ undefined. Correct move: Apply domain restrictions to every logarithmic term in the original equation, not just the combined one.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following is the solution set to the equation $\ln (x^2-4)=\ln (3x)$? A) $\{4\}$ B) $\{-1,4\}$ C) $\{-1\}$ D) No real solutions

Worked Solution: First, find the domain of the original equation. For $\ln (x^2-4)$, we need $x^2-4>0⟹|x|>2$. For $\ln (3x)$, we need $3x>0⟹x>0$. The combined domain is $x>2$. Apply the one-to-one property of logarithms to get $x^2-4=3x$, which rearranges to $x^2-3x-4=0$. Factoring gives $(x-4)(x+1)=0$, so solutions are $x=4$ and $x=-1$. Checking against the domain, $x=-1<2$ is extraneous, so only $x=4$ is valid. The correct answer is A.

Question 2 (Free Response)

Consider the function $f(x)=2e^{0.1x}-5$. (a) Solve $f(x)=3$ for $x$, giving both exact and approximate value to 3 decimal places. (b) Find all solutions to $f(x)>3$, writing your answer in interval notation. (c) Find the domain of $g(x)=\ln (f(x))$, and write it in interval notation.

Worked Solution: (a) Start with $2e^{0.1x}-5=3$. Add 5 to both sides: $2e^{0.1x}=8$. Divide by 2: $e^{0.1x}=4$. Take the natural logarithm of both sides: $0.1x=\ln 4$. Solve for $x$: $x=10\ln 4\approx 13.863$. Exact: $x=10\ln 4$, approximate: $x\approx 13.863$.

(b) Start with $2e^{0.1x}-5>3$, which simplifies to $e^{0.1x}>4$. Since $e>1$, the exponential function is increasing, so inequality direction is preserved: $0.1x>\ln 4⟹x>10\ln 4\approx 13.863$. In interval notation: $(10\ln 4,\infty )\approx (13.863,\infty )$.

(c) The domain of $\ln (f(x))$ requires $f(x)>0$, so $2e^{0.1x}-5>0⟹e^{0.1x}>2.5⟹0.1x>\ln 2.5⟹x>10\ln 2.5\approx 9.163$. The domain is $(10\ln 2.5,\infty )\approx (9.163,\infty )$.

Question 3 (Application / Real-World Style)

A biologist is studying bacterial growth in a petri dish. The number of bacteria $N$ (in thousands) $t$ hours after the start of the experiment is given by $N(t)=1.5e^{0.25t}$, where $t\ge 0$. (a) How many hours will it take for the number of bacteria to reach 12 thousand? Round your answer to the nearest tenth of an hour. (b) The biologist will stop the experiment once the number of bacteria exceeds 50 thousand. What is the range of times when the experiment is still running?

Worked Solution: (a) We solve $1.5e^{0.25t}=12$. Divide by 1.5: $e^{0.25t}=8$. Take natural logs: $0.25t=\ln 8$. Solve for $t$: $t=\frac{\ln 8}{0.25}\approx \frac{2.0794}{0.25}\approx 8.3$ hours. In context, it takes approximately 8.3 hours for the bacteria population to reach 12 thousand.

(b) We need $N(t)\le 50$, so $1.5e^{0.25t}\le 50⟹e^{0.25t}\le \frac{50}{1.5}\approx 33.33$. Take logs: $0.25t\le \ln (33.33)\approx 3.507$, so $t\le \frac{3.507}{0.25}\approx 14.0$ hours. The experiment starts at $t=0$, so it is still running for times $0\le t\le 14.0$ hours, or $[0,14.0]$ hours.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the core skill for all applications of exponential and logarithmic functions, which make up 28-35% of the AP Precalculus exam. Immediately next, you will apply these solving skills to model exponential growth and decay, half-life, compound interest, and logistic growth in contextual problems. Without mastering the ability to correctly solve these equations and inequalities—including checking for extraneous solutions and correctly handling inequality direction—you will not be able to correctly interpret real-world models or answer application-based FRQs. This topic also builds the foundation for future calculus study, where you will differentiate and integrate exponential and logarithmic functions, requiring you to solve for unknown constants using these same core skills.

Exponential and logarithmic function properties Modeling with exponential and logarithmic functions Logarithmic and exponential function graphs