Logarithmic functions — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Definition of logarithmic functions, inverse relationship with exponential functions, core logarithm properties, change of base formula, solving logarithmic equations, and graph transformations of logarithmic functions.

You should already know: Exponential function properties, inverse function definitions, basic exponent algebra.

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 functions?

A logarithmic function is the inverse of a one-to-one exponential function, defined formally by the relationship: for $b>0,b\ne 1$, $y={\log }_{b}x$ if and only if $b^y=x$. Standard notation conventions reserve $\log x$ for the common logarithm (base 10) and $\ln x$ for the natural logarithm (base $e\approx 2.71828$), the most commonly used form in calculus and continuous growth models. According to the AP Precalculus CED, Unit 2 (Exponential and Logarithmic Functions) accounts for 25-35% of total exam weight, and logarithmic function content makes up roughly half of that unit, appearing in both multiple-choice (MCQ) and free-response (FRQ) sections. Logarithmic functions reverse exponential operations, allowing us to solve for unknown exponents and model phenomena that scale logarithmically, from pH levels in chemistry to decibel ratings for sound to doubling time in population growth. Unlike exponential functions, which have an unrestricted domain and positive range, logarithmic functions have a domain of only positive real numbers ($x>0$) and an unrestricted range of all real numbers, a direct result of their inverse relationship to exponentials.

2. Core Logarithm Properties and Change of Base

Since logarithmic functions are inverses of exponential functions, all logarithm properties are derived directly from corresponding exponent rules. The key inverse identities that connect the two function types are: $$b^{{\log }_{b}x}=x(x>0)\text{and}{\log }_b(b^x)=x(\text{for all real }x)$$ From exponent rules, we get the three core logarithm properties:

1. Product Rule: ${\log }_b(MN)={\log }_{b}M+{\log }_{b}N$ (exponents add when multiplying powers, so logs turn products into sums)
2. Quotient Rule: ${\log }_b\left(\frac{M}{N}\right)={\log }_{b}M-{\log }_{b}N$ (exponents subtract when dividing powers)
3. Power Rule: ${\log }_b(M^k)=k{\log }_{b}M$ (exponents multiply when raising a power to a power) The change of base formula allows us to evaluate any logarithm with a standard calculator, which only computes base 10 or base $e$ logarithms: $${\log }_{b}a=\frac{{\log }_{c}a}{{\log }_{c}b}(c>0,c\ne 1,a>0,b>0,b\ne 1)$$ Most often, we use $c=10$ or $c=e$, so ${\log }_{b}a=\frac{\ln a}{\ln b}=\frac{\log a}{\log b}$.

Worked Example

Problem: Simplify ${\log }_5(250)-{\log }_5(2)+3{\log }_5(\sqrt{5})$ to a simplified numerical value.

1. Apply the quotient logarithm rule to the first two terms: ${\log }_5\left(\frac{250}{2}\right)+3{\log }_5(\sqrt{5})={\log }_5(125)+3{\log }_5(5^{1/2})$.
2. Apply the power rule to the second term: $3{\log }_5(5^{1/2})=3\cdot \frac{1}{2}\cdot {\log }_{5}5=\frac{3}{2}(1)=\frac{3}{2}$, since ${\log }_{b}b=1$ for any valid base $b$.
3. Rewrite 125 as a power of 5: $125=5^3$, so ${\log }_5(5^3)=3{\log }_{5}5=3(1)=3$ by the inverse identity.
4. Add the results: $3+\frac{3}{2}=\frac{9}{2}=4.5$. Final result: $\frac{9}{2}$ (or $4.5$).

Exam tip: Always check the domain of every argument in a logarithmic expression before simplifying—even if your algebra gives a nice number, if any argument is non-positive, the expression is undefined. AP MCQ distractors often include the simplified numeric answer ignoring domain checks.

3. Solving Logarithmic Equations

A logarithmic equation is any equation where the unknown variable appears inside the argument of a logarithm. The core strategy for solving these equations leverages the inverse relationship between logs and exponentials: first isolate a single logarithmic term on one side of the equation, then convert the equation to exponential form to eliminate the logarithm. If you have multiple logarithms on the same side, use logarithm properties to combine them into a single term before converting. The most critical step that many students miss is checking for extraneous solutions: since logarithms only have positive arguments, any solution that makes any original logarithm's argument zero or negative must be discarded, even if it satisfies the final equation after combining logs. Combining multiple logs can erase individual domain restrictions, so always check against the original equation.

Worked Example

Problem: Solve ${\log }_2(x+2)+{\log }_2(x)=3$ for all real solutions.

1. First write domain restrictions from the original equation: $x+2>0⟹x>-2$ and $x>0$, so the overall valid domain is $x>0$; any candidate solution $\le 0$ is automatically invalid.
2. Combine the two logarithms with the product rule: ${\log }_2\left(x(x+2)\right)=3⟹{\log }_2(x^2+2x)=3$.
3. Convert to exponential form by the definition of logarithms: $2^3=x^2+2x⟹x^2+2x-8=0$.
4. Factor the quadratic: $(x+4)(x-2)=0$, giving candidate solutions $x=-4$ and $x=2$.
5. Check against the domain: $x=-4<0$ is extraneous and discarded, while $x=2>0$ is valid. Final solution: $x=2$.

Exam tip: Always check solutions against the domain of the original equation, not just the final equation after combining logs. Extraneous solutions are a common feature of AP logarithmic equation questions, and distractors almost always include the extraneous solution.

4. Graphing and Transformations of Logarithmic Functions

The parent logarithmic function $y={\log }_{b}x$ has consistent core features: it has a vertical asymptote at $x=0$ (the y-axis), passes through the point $(1,0)$ (since ${\log }_{b}1=0$ for any valid $b$), has domain $(0,\infty )$ and range all real numbers. If $b>1$, the function is strictly increasing and concave down, approaching $-\infty$ as $x\to 0^{+}$. If $0<b<1$, the function is strictly decreasing and concave up, approaching $+\infty$ as $x\to 0^{+}$. Transformations of logarithmic functions follow the same rules as all other function transformations: for $y=a{\log }_b(c(x-h))+k$, $a$ controls vertical stretch/compression/reflection, $c$ controls horizontal stretch/compression/reflection, $h$ is the horizontal shift, and $k$ is the vertical shift. The vertical asymptote shifts with the horizontal shift: to find the new asymptote, set the argument of the logarithm equal to zero and solve for $x$, which is simpler than memorizing shift directions.

Worked Example

Problem: For $f(x)=\ln (x-3)+2$, identify the domain, equation of the vertical asymptote, and one point on the graph, then describe the transformation from the parent $y=\ln x$.

1. Find the domain by requiring the argument to be positive: $x-3>0⟹x>3$, so domain is $(3,\infty )$ in interval notation.
2. Find the vertical asymptote by setting the argument equal to zero: $x-3=0⟹x=3$, which is the equation of the vertical asymptote.
3. To find a simple point on the graph, use the parent function's known intercept: when $\ln (\text{argument})=0$, the argument equals 1. Set $x-3=1⟹x=4$, so $f(4)=\ln (1)+2=0+2=2$, so $(4,2)$ is a point on the graph.
4. Describe the transformation: the graph of $y=\ln x$ is shifted 3 units right and 2 units up to produce $f(x)$.

Exam tip: When asked for the vertical asymptote of a logarithmic function on an FRQ, always write your answer as an equation ($x=h$, not just $h$). AP graders regularly deduct points for missing the "x=" part of the answer.

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 with adding arguments, misremembering that logs turn products into sums, not sums into sums. Correct move: Always recall that only ${\log }_b(MN)={\log }_{b}M+{\log }_{b}N$; there is no general simplification for the logarithm of a sum.
• Wrong move: After solving ${\log }_2(x+2)+{\log }_{2}x=3$, keeping $x=-4$ as a valid solution. Why: Students check solutions against the final combined equation, not the original equation, forgetting that individual terms have domain restrictions lost when combining. Correct move: Always list domain restrictions from the original equation before solving, and discard any candidate that violates any restriction.
• Wrong move: $\frac{{\log }_{b}M}{{\log }_{b}N}={\log }_b\left(\frac{M}{N}\right)$. Why: Students confuse the change of base formula with the quotient rule, mixing up the order of operations. Correct move: Remember the quotient rule is ${\log }_b\left(\frac{M}{N}\right)={\log }_{b}M-{\log }_{b}N$, while $\frac{{\log }_{b}M}{{\log }_{b}N}={\log }_{N}M$ by change of base—they are not equal.
• Wrong move: Stating the domain of $f(x)=\log (x^2)$ is $x>0$. Why: Students generalize the domain rule for $\log (x)$ to $\log (x^2)$ without checking when the argument is positive. Correct move: Always check for positive arguments directly: $x^2>0$ for all $x\ne 0$, so domain is $(-\infty ,0)\cup (0,\infty )$.
• Wrong move: Writing the vertical asymptote of $y=\ln (x+2)-1$ as $x=2$. Why: Students mix up horizontal shift direction, assuming $+2$ shifts right instead of left. Correct move: Always find the vertical asymptote by setting the argument equal to zero and solving: $x+2=0⟹x=-2$, which gives the correct asymptote regardless of shift direction.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following is equivalent to $3\ln 2-\ln 4$? A) $\ln 2$ B) $\ln 6$ C) $2\ln 3$ D) $-\ln 2$

Worked Solution: First apply the power rule of logarithms to rewrite $3\ln 2=\ln (2^3)=\ln 8$. Next, apply the quotient rule: $\ln 8-\ln 4=\ln \left(\frac{8}{4}\right)=\ln 2$. Option B comes from incorrectly adding 2 and 3 instead of raising 2 to the 3rd power, option D incorrectly rewrites $3\ln 2$ as $\ln 2^{-3}$, and option C is an unrelated distractor. The correct answer is A.

Question 2 (Free Response)

Let $f(x)={\log }_3(2x-3)$. (a) State the domain of $f(x)$ and find the equation of its vertical asymptote. (b) Solve $f(x)=2$ for $x$. (c) Find $f^{-1}(x)$, the inverse function of $f(x)$.

Worked Solution: (a) The argument of a logarithm must be positive, so $2x-3>0⟹x>\frac{3}{2}$. Domain is $\left(\frac{3}{2},\infty \right)$. The vertical asymptote occurs where the argument equals zero, so $2x-3=0⟹x=\frac{3}{2}$ is the vertical asymptote. (b) Substitute $f(x)=2$: ${\log }_3(2x-3)=2$. Rewrite in exponential form: $3^2=2x-3⟹9=2x-3⟹12=2x⟹x=6$. Check that $2(6)-3=9>0$, so $x=6$ is valid. (c) To find the inverse, swap $x$ and $y$: $x={\log }_3(2y-3)$. Rewrite in exponential form: $3^x=2y-3$. Solve for $y$: $2y=3^x+3⟹y=\frac{3^x+3}{2}$. So $f^{-1}(x)=\frac{3^x+3}{2}$.

Question 3 (Application / Real-World Style)

The pH of an aqueous solution is defined as $\text{pH}=-{\log }_{10}[H^{+}]$, where $[H^{+}]$ is the concentration of hydrogen ions in moles per liter (mol/L). A sample of orange juice has a pH of 3.8, and a sample of black coffee has a pH of 5.0. (a) What is the hydrogen ion concentration of the orange juice, rounded to 3 significant figures? (b) How many times greater is the hydrogen ion concentration in orange juice than in black coffee?

Worked Solution: (a) Substitute pH = 3.8 into the formula: $3.8=-{\log }_{10}[H^{+}]⟹{\log }_{10}[H^{+}]=-3.8$. Rewrite in exponential form: $[H^{+}]=10^{-3.8}\approx 1.58\times 10^{-4}$ mol/L. (b) First find the hydrogen ion concentration of black coffee: $5.0=-{\log }_{10}[H^{+}{]}_{\text{coffee}}⟹[H^{+}{]}_{\text{coffee}}=10^{-5.0}=1.0\times 10^{-5}$ mol/L. Calculate the ratio of concentrations: $\frac{1.58\times 10^{-4}}{1.0\times 10^{-5}}=15.8\approx 16$. In context, the hydrogen ion concentration of orange juice is approximately 16 times greater than that of black coffee, which confirms that orange juice is significantly more acidic than black coffee, consistent with its lower pH.

7. Quick Reference Cheatsheet

8. What's Next

Mastery of logarithmic functions is required for all remaining topics in Unit 2 of AP Precalculus, starting with exponential and logarithmic modeling, where you will use logarithms to solve for unknown parameters in continuous growth and decay models. Without solid proficiency in simplifying logarithmic expressions and checking for extraneous solutions when solving logarithmic equations, you will not be able to calculate half-life, doubling time, or carry out parameter estimation for exponential data, a frequent free-response question topic on the AP exam. Logarithmic functions also lay critical groundwork for future calculus study, where the natural logarithm is the antiderivative of $\frac{1}{x}$, and logarithmic differentiation simplifies complex derivative problems. Next you will apply logarithmic properties to model real-world exponential phenomena: Exponential growth and decay modeling Solving exponential equations Logarithmic regression