# Exponential and logarithmic equations and inequalities

> AP Precalculus · Unit 2: Exponential and Logarithmic Functions
> Source: https://www.owlsprep.com/study/ap-precalculus-u2-exponential-and-logarithmic-equations-and/

This module covers solving exponential and logarithmic equations and inequalities, using one-to-one properties and inverse function relationships. You will learn to identify extraneous solutions and correctly handle inequality direction for AP Precalculus.

**Prerequisites:** [Properties of exponents and logarithms](https://www.owlsprep.com/study/ap-precalculus-u2-exponential-logarithm-properties/); Solving linear and quadratic equations; Domain restrictions for logarithmic functions

## Learning objectives

- Solve exponential equations using common base and logarithm methods
- Solve logarithmic equations and identify extraneous solutions
- Solve exponential and logarithmic inequalities accounting for function monotonicity
- Apply solving techniques to real-world modeling problems

## Solving Exponential Equations

An exponential equation has the general form $a^{f(x)} = b^{g(x)}$, where $a,b>0$ and $a,b \neq 1$. There are two core solution methods, depending on whether the bases can be rewritten to match.

If both sides can be rewritten with the same base, apply the one-to-one property to drop the base and solve the resulting polynomial equation directly. If bases cannot be matched, take the natural or common logarithm of both sides, apply the power rule $\ln a^b = b\ln a$ to bring the exponent down, then solve for $x$.

**General Exponential Solution** — For equations where bases cannot be matched, the exact solution is derived by taking logarithms of both sides

*Notation:* $a^{kx + c} = d$

*Example:* $x = \frac{\ln d - c\ln a}{k\ln a}$

**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 for $x$:

   $$4x\ln 2 + 3\ln 2 = x\ln 5 - 2\ln 5 \implies x(4\ln 2 - \ln 5) = -2\ln 5 - 3\ln 2$$
5. Solve for $x$ to get exact and approximate forms:

   $$x = \frac{-(2\ln 5 + 3\ln 2)}{4\ln 2 - \ln 5} \approx -4.553$$

**Exam command terms**

- **Give your answer to three decimal places** — You must provide an approximate decimal value, not just an exact form *(AP Precalc expects 3 decimal places for all approximate answers unless stated otherwise)*

> **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.

## Solving Logarithmic Equations

To solve logarithmic equations, we use the one-to-one property for matching logarithms or rewrite the equation in exponential form using the definition of a logarithm. The most critical step is checking for extraneous solutions, which occur when a solution makes the argument of any original logarithm non-positive.

**Extraneous Solution** — A solution that satisfies the rewritten algebraic equation but does not satisfy the original equation's domain restrictions. All solutions to logarithmic equations must be checked against the original domain.

*Example:* $x=-3$ is extraneous for $\log_2(x+2) + \log_2(x-1) = 2$

**Worked example:** Solve $\log_2 (x+2) + \log_2 (x-1) = 2$.

1. First, write domain restrictions from the original equation: $x+2>0 \implies x>-2$, and $x-1>0 \implies x>1$, so the valid 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 standard quadratic form:

   $$x^2 + x - 2 = 4 \implies x^2 + x - 6 = 0$$
5. Factor and solve the quadratic:

   $$(x+3)(x-2) = 0 \implies x = -3 \text{ or } x=2$$
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.

## Solving Exponential and Logarithmic Inequalities

Exponential and logarithmic inequalities follow the same initial steps as equations: first find the domain, then rewrite to use the one-to-one property. The key difference is adjusting inequality direction based on whether the function is increasing or decreasing.

For base $b>1$, both $b^x$ and $\log_b x$ are strictly increasing, so inequality direction is preserved when dropping the base or logarithm. For $0<b<1$, both functions are strictly decreasing, so inequality direction is reversed when dropping the base or logarithm. The final solution set is the intersection of the simplified solution with the original domain.

**Worked example:** Solve $\log_{0.5}(2x - 3) < \log_{0.5}(x + 1)$.

1. Find the domain of the original inequality: $2x-3>0 \implies x>1.5$, and $x+1>0 \implies x>-1$, 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. We reverse the inequality sign when dropping the logarithm.
3. Rewrite the simplified 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.

## AP-Style Practice Problems

**Check your understanding**

Test your understanding with these AP-style problems

1. 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

   *Why:* First find the domain: $x^2-4>0$ gives $|x|>2$, and $3x>0$ gives $x>0$, so combined domain is $x>2$. Applying the one-to-one property gives $x^2-3x-4=0$, with roots $x=4$ and $x=-1$. Only $x=4$ is in the domain, so it is the only solution.

## Common pitfalls

- **Wrong:** When solving $\log_b A = \log_b C$, immediately conclude $A=C$ and keep all solutions without checking the original domain.
  - Why it fails: Students remember the one-to-one property but forget that only positive arguments are valid, so extraneous solutions are often left in.
  - Correct: 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:** When solving any exponential inequality, automatically reverse the inequality sign regardless of the base value.
  - Why it fails: Students confuse the base rule for inequality direction, reversing the sign when it should be preserved for bases greater than 1.
  - Correct: Before dropping the base, explicitly check the base: if $b>1$, keep inequality direction; if $0<b<1$, reverse it.
- **Wrong:** When solving $5^{2x} = 3^x$, after expanding to $2x \ln 5 = x \ln 3$, divide both sides by $x$, losing the solution $x=0$.
  - Why it fails: Dividing by a variable expression assumes it is non-zero, which eliminates any potential root at $x=0$.
  - Correct: Factor out the common variable term instead of dividing: $x(2 \ln 5 - \ln 3) = 0$, which captures all solutions.
- **Wrong:** When rewriting $\ln (2x + 1)^2 = 6$, apply the power rule to get $2 \ln (2x + 1) = 6$, leading to only one solution.
  - Why it fails: The power rule $\ln a^b = b \ln a$ only holds when $a>0$; squaring makes the argument positive even if $2x+1$ is negative, so solutions can be lost.
  - Correct: Rewrite in exponential form first: $(2x+1)^2 = e^6$, so $2x+1 = \pm e^3$, giving both valid solutions.
- **Wrong:** When solving a sum of logarithms, after combining into a single logarithm, only check that the combined argument is positive.
  - Why it fails: The combined argument can be positive for values that make an original individual logarithm's argument negative, leading to an invalid solution.
  - Correct: Apply domain restrictions to every logarithmic term in the original equation, not just the combined one.

## Cheatsheet

| Category | Rule / Formula | Notes |
| --- | --- | --- |
| One-to-One Property (Exponentials) | If $a^m = a^n$, $a>0, a\neq1$, then $m=n$ | Only applies when bases are equal; works for any real exponents |
| General Solution (Exponential Equations) | $a^{kx + c} = d \implies x = \frac{\ln d - c \ln a}{k \ln a}$ | Use when bases cannot be rewritten to a common base |
| One-to-One Property (Logarithms) | If $\log_b A = \log_b C$, $b>0, b\neq1$, then $A=C$ | Requires $A>0, C>0$; always check for extraneous solutions |
| Log to Exponential Conversion | $\log_b A = k \iff b^k = A$ | Requires $A>0$, $b>0$, $b\neq1$ |
| Inequality Rule: $b>1$ | $b^f < b^g \implies f < g$; $\log_b f < \log_b g \implies f < g$ | Both functions are increasing, so inequality direction is preserved |
| Inequality Rule: $0<b<1$ | $b^f < b^g \implies f > g$; $\log_b f < \log_b g \implies f > g$ | Both functions are decreasing, so inequality direction is reversed |
| Logarithm Power Rule | $\ln a^b = b \ln a$ | Only valid when $a>0$; for $\ln a^2$, use $2 \ln \|a\|$ to preserve all solutions |
| Domain Rule for Log Equations | All arguments of all logarithms must be strictly positive | Apply to original equation, not just combined logarithms |

## What's next

This topic is the core skill for all applications of exponential and logarithmic functions, which make up 28-35% of the total AP Precalculus exam. Mastering these solving techniques is essential for correctly answering both multiple-choice and free-response questions, especially application-based problems that rely on finding unknown values in real-world models. Immediately next, you will apply these skills to model exponential growth and decay, half-life, and compound interest in contextual problems. Without correctly solving equations, checking for extraneous solutions, and handling inequality direction correctly, you will not be able to correctly interpret models or earn full points on FRQs. This topic also builds the foundation for future calculus study, where you will work extensively with exponential and logarithmic functions.

- [Logarithmic function context and data modeling](https://www.owlsprep.com/study/ap-precalculus-u2-logarithmic-function-context-and-data/)
- [Semi-log Plots](https://www.owlsprep.com/study/ap-precalculus-u2-semi-log-plots/)
- [Trigonometric and Polar Functions Overview](https://www.owlsprep.com/study/ap-precalculus-u3-overview/)

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