Logarithmic expressions — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Definition of logarithmic expressions, logarithm-exponential equivalence, product/quotient/power logarithm rules, change of base formula, and techniques for expanding and condensing logarithmic expressions for algebraic manipulation.

You should already know: Exponential function definitions and properties; algebraic rules for exponents; basic factoring and simplification of algebraic expressions.

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 Logarithmic expressions?

A logarithmic expression is any algebraic expression containing one or more logarithms, which are the inverse functions of exponential functions. Per the AP Precalculus Course and Exam Description (CED), this topic is part of Unit 2: Exponential and Logarithmic Functions, which accounts for 25-30% of the total AP exam score. Logarithmic expressions appear in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam, most often as a foundational step for solving exponential/logarithmic equations, modeling real-world growth and decay, and simplifying functional expressions.

Unlike exponential expressions, which raise a base to an exponent to get a result, logarithms answer the question: what exponent do I raise this base to, to get the given result? On the AP exam, you will primarily work with common logarithms (base 10, written $\log x$) and natural logarithms (base $e$, written $\ln x$), though any positive base $b\ne 1$ is valid. Mastery of simplifying logarithmic expressions is non-negotiable for all subsequent Unit 2 topics, so errors here will cascade into mistakes on higher-weight exam problems.

2. Logarithm-Exponential Equivalence

The entire definition of a logarithm rests on its inverse relationship with exponentiation. For any real exponent $y$, positive base $b\ne 1$, and positive argument $x$, the core equivalence is: $$y={\log }_{b}x⟺b^y=x$$ This equivalence lets you convert freely between exponential and logarithmic forms, which is the starting point for every problem involving logarithms. Key domain restrictions are non-negotiable for the AP exam: the base $b$ must always be positive and not equal to 1 (a base of 1 is constant and cannot produce variable outputs, while a negative base leads to non-real outputs for most exponents). The argument of a logarithm (the $x$ in ${\log }_{b}x$) must always be positive—you cannot get a negative output by raising a positive base to any real exponent, so negative arguments produce non-real logarithms. This equivalence is how you evaluate simple logarithmic expressions by hand, a common skill on the no-calculator MCQ section.

Worked Example

Evaluate ${\log }_4\left(\frac{1}{64}\right)$.

1. Let $y={\log }_4\left(\frac{1}{64}\right)$. By the logarithm-exponential equivalence, this translates to $4^y=\frac{1}{64}$.
2. Rewrite both sides with a common base of 4 to match the logarithm's base: $\frac{1}{64}=64^{-1}=(4^3{)}^{-1}=4^{-3}$, so we get $4^y=4^{-3}$.
3. Since exponential functions are one-to-one, their exponents must be equal: $y=-3$.
4. Verify the result: $4^{-3}=\frac{1}{4^3}=\frac{1}{64}$, which matches the original argument.

Exam tip: If you are ever unsure of a logarithm value, just convert it back to exponential form to check—this takes 10 seconds and eliminates almost all sign errors.

3. Core Logarithm Properties (Product, Quotient, Power Rules)

Because logarithms are inverses of exponents, their properties directly correspond to familiar exponent rules. For any positive base $b\ne 1$, positive arguments $M,N$, and real number $k$, the three core properties are:

1. Product Rule: ${\log }_b(MN)={\log }_{b}M+{\log }_{b}N$ (corresponds to $b^m{b}^n=b^{m+n}$)
2. Quotient Rule: ${\log }_b\left(\frac{M}{N}\right)={\log }_{b}M-{\log }_{b}N$ (corresponds to $\frac{b^m}{b^n}=b^{m-n}$)
3. Power Rule: ${\log }_b(M^k)=k{\log }_{b}M$ (corresponds to $(b^m{)}^k=b^{mk}$)

These rules are used for two high-frequency tasks on the AP exam: expanding a single condensed logarithmic expression into a sum of simpler terms, and condensing a sum/difference of logarithms into a single logarithmic expression. Both tasks are routine steps in solving logarithmic equations and simplifying expressions for later calculus work.

Worked Example

Expand ${\log }_3\left(\frac{9x^5}{\sqrt{y}}\right)$ fully, where $x>0,y>0$.

1. Apply the quotient rule first to split the fraction: ${\log }_3\left(\frac{9x^5}{\sqrt{y}}\right)={\log }_3(9x^5)-{\log }_3(\sqrt{y})$.
2. Apply the product rule to the first term, then rewrite the square root as an exponent: ${\log }_{3}9+{\log }_3(x^5)-{\log }_3(y^{1/2})$.
3. Evaluate the constant log term and apply the power rule to the variable terms: ${\log }_{3}9=2$, so we get $2+5{\log }_{3}x-\frac{1}{2}{\log }_{3}y$.
4. This is fully expanded: no products, quotients, or powers remain inside any logarithm.

Exam tip: When condensing expressions, always move coefficients inside logarithms as exponents before combining terms with product/quotient rules—this avoids common errors where coefficients are incorrectly applied to the combined term.

4. Change of Base Formula

The change of base formula lets you rewrite a logarithm of any base as a ratio of logarithms with any other base, which is required for evaluating logarithms with a calculator and simplifying expressions with mixed bases. The formula states that for any positive $a,b\ne 1$, and positive $x$: $${\log }_{b}x=\frac{{\log }_{a}x}{{\log }_{a}b}$$ On the AP exam, the two most common uses are converting to base 10 or base $e$ to evaluate with a calculator (most calculators only have $\log$ and $\ln$ buttons) and simplifying expressions with related bases (e.g., simplifying ${\log }_{4}x$ to $\frac{1}{2}{\log }_{2}x$). The formula is easily derived from the core equivalence: let $y={\log }_{b}x$, so $b^y=x$. Take ${\log }_a$ of both sides, apply the power rule to get $y{\log }_{a}b={\log }_{a}x$, then solve for $y$ to get the formula.

Worked Example

Simplify $\frac{{\log }_{2}27}{{\log }_{2}9}$ to a single constant.

1. Recognize that the expression matches the right-hand side of the change of base formula, with $a=2$, $b=9$, $x=27$. This means the expression simplifies directly to ${\log }_{9}27$.
2. Rewrite 27 and 9 as powers of the same base 3: ${\log }_{3^2}{3}^3$.
3. Apply the change of base formula again to rewrite in base 3: $\frac{{\log }_3{3}^3}{{\log }_3{3}^2}=\frac{3{\log }_{3}3}{2{\log }_{3}3}$.
4. Cancel the common ${\log }_{3}3=1$ term to get $\frac{3}{2}$, which can be verified by checking $9^{3/2}=27$.

Exam tip: When asked to evaluate a logarithm of a number that is a power of the logarithm's base power, always rewrite both terms with the same small base to cancel out logs and avoid calculator work.

5. Common Pitfalls (and how to avoid them)

• Wrong move: ${\log }_b(M+N)={\log }_{b}M+{\log }_{b}N$. Why: Students confuse the product rule for logs with addition inside the log; there is no general rule for the logarithm of a sum. Correct move: Always check what operation is inside the log before applying a rule—only product, quotient, and power can be split, not addition or subtraction.
• Wrong move: $\frac{{\log }_{b}M}{{\log }_{b}N}={\log }_b\left(\frac{M}{N}\right)$. Why: Students confuse the change of base formula (which describes a quotient of two logs) with the quotient rule (which applies to a quotient inside a single log). Correct move: Only apply the quotient rule when the division is inside the logarithm, not when it is between two separate logarithms.
• Wrong move: ${\log }_b(-x)=-{\log }_{b}x$ for $x>0$. Why: Students incorrectly apply the power rule to a negative argument, forgetting the domain requirement that arguments must be positive. Correct move: Always check the domain first: ${\log }_b(-x)$ is only defined for $x<0$, and you can never pull a negative sign out of an argument as a coefficient.
• Wrong move: $({\log }_{b}M{)}^k=k{\log }_{b}M$. Why: Students misapply the power rule, which only applies to a power inside the logarithm, not to the entire logarithm raised to a power. Correct move: Only use the power rule when the exponent is on the argument of the log, not on the log itself.
• Wrong move: ${\log }_b\left(\frac{1}{x}\right)=\frac{1}{{\log }_{b}x}$. Why: Students incorrectly apply reciprocal rules from general algebra to logarithms, forgetting the power rule for reciprocals. Correct move: Rewrite $\frac{1}{x}$ as $x^{-1}$, then apply the power rule to get ${\log }_b\left(\frac{1}{x}\right)=-{\log }_{b}x$.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following is equivalent to $2\log 6-\log 3$? A) $\log 3$ B) $\log 8$ C) $\log 12$ D) $\log 36$

Worked Solution: First, apply the power rule to move the coefficient of the first term to be an exponent on the argument: $2\log 6=\log 6^2=\log 36$. Next, apply the quotient rule to combine the two logarithms through subtraction: $\log 36-\log 3=\log \left(\frac{36}{3}\right)=\log 12$. The other options come from common errors: A reverses the order of subtraction, B incorrectly adds coefficients instead of applying exponent rules, D forgets to apply the quotient rule. The correct answer is C.

Question 2 (Free Response)

Let $x>0,y>0$, and let $A={\log }_2(16x^3{y}^{-2})$. (a) Expand $A$ fully into a sum of constant and simple logarithmic terms with no powers or products inside any logarithm. (b) Given that ${\log }_{2}x=4$ and ${\log }_{2}y=-1$, calculate the numerical value of $A$. (c) Rewrite $A$ in terms of $\ln x$ and $\ln y$.

Worked Solution: (a) First apply the product rule to split the argument: $${\log }_{2}16+{\log }_2{x}^3+{\log }_2{y}^{-2}$$ Apply the power rule to variable terms, then evaluate the constant log: $${\log }_{2}16=4⟹A=4+3{\log }_{2}x-2{\log }_{2}y$$

(b) Substitute the given values into the expanded expression: $$A=4+3(4)-2(-1)=4+12+2=18$$

(c) Use the change of base formula to convert base 2 logs to natural logs: ${\log }_{2}x=\frac{\ln x}{\ln 2}$ and ${\log }_{2}y=\frac{\ln y}{\ln 2}$. Substitute into the expanded form: $$A=4+3\left(\frac{\ln x}{\ln 2}\right)-2\left(\frac{\ln y}{\ln 2}\right)=4+\frac{3\ln x-2\ln y}{\ln 2}$$

Question 3 (Application / Real-World Style)

The pH of a liquid solution is given by the formula $\text{pH}=-{\log }_{10}[{\text{H}}^{+}]$, where $[{\text{H}}^{+}]$ is the concentration of hydrogen ions in moles per liter (mol/L). Orange juice has a hydrogen ion concentration of $3.2\times 10^{-4}$ mol/L. Calculate the pH of orange juice to one decimal place. Milk has a pH of 6.6. What is the hydrogen ion concentration of milk, expressed in scientific notation?

Worked Solution: First, substitute the orange juice hydrogen ion concentration into the pH formula: $$\text{pH}=-{\log }_{10}(3.2\times 10^{-4})$$ Use the product rule to expand, then simplify: $$\text{pH}=-\left({\log }_{10}3.2+{\log }_{10}10^{-4}\right)=-\left(0.505-4\right)=3.495\approx 3.5$$

For milk, rearrange the pH formula and convert to exponential form: $$6.6=-{\log }_{10}[{\text{H}}^{+}]⟹-6.6={\log }_{10}[{\text{H}}^{+}]⟹[{\text{H}}^{+}]=10^{-6.6}=10^{0.4}\times 10^{-7}\approx 2.5\times 10^{-7}$$

In context: Orange juice has a pH of 3.5, which is acidic (consistent with its taste), and milk has a lower hydrogen ion concentration than orange juice, matching its higher near-neutral pH.

7. Quick Reference Cheatsheet

8. What's Next

Mastery of logarithmic expressions is the foundational prerequisite for all remaining topics in Unit 2: Exponential and Logarithmic Functions. Next, you will apply these simplification and manipulation techniques to solving exponential and logarithmic equations, where you will need to expand or condense expressions to isolate the variable of interest. Without being able to correctly apply logarithm properties and convert between exponential and logarithmic forms, you will not be able to correctly solve these equations, which make up a large portion of the unit's exam weight. This topic also feeds into modeling real-world exponential growth and decay, where logarithmic expressions are used to calculate the time required to reach a given threshold value, and for graphing logarithmic functions.

solving exponential and logarithmic equations modeling with exponential and logarithmic functions graphs of logarithmic functions