AP® Precalculus review sheet from Aim for Five (aimforfive.com/precalc/units/2/2-11)
Unit 2 · Topic 2.11
2.11 Logarithmic Functions
A logarithmic function only takes positive inputs, has a vertical asymptote at x = 0, and grows without bound, but extremely slowly. Like exponential functions, logs always rise or always fall and always bend the same way.
Key terms
- domain of a log
- vertical asymptote
- concavity
- logarithmic growth
Domain, range and asymptote
For f(x) = a · log_b(x) in general form, the domain is x > 0 and the range is all real numbers.
The graph has a vertical asymptote at x = 0. Near it, the outputs blow up. For log₂(x): lim (x→0⁺) log₂(x) = −∞. At the other end, lim (x→∞) log₂(x) = ∞, so the outputs are unbounded both ways even though they grow slowly.
Always one direction, always one concavity
A log graph is an exponential graph reflected over the line y = x, so it inherits the same steadiness: it never turns around and never changes concavity. So a log function has no turning points and no points of inflection, and it has extrema only on a closed interval.
| Function | Direction | Concavity | lim (x→0⁺) | lim (x→∞) |
|---|---|---|---|---|
| a > 0, b > 1 (like log₂ x) | Increasing | Concave down | −∞ | ∞ |
| a < 0, b > 1 (like −log₂ x) | Decreasing | Concave up | ∞ | −∞ |
| a > 0, 0 < b < 1 (like log₀.₅ x) | Decreasing | Concave up | ∞ | −∞ |
Very slow growth
log₂(x) reaches 10 at x = 1,024 and reaches 20 only at x = 1,048,576. Doubling the input adds just 1 to the output.
Compare exponential growth: 2ˣ goes from 1,024 to over a million as x goes from 10 to 20. Logs undo that growth, so they are the slowest-growing of the common unbounded functions in this course.
Slow is not the same as bounded. For any target output, however large, some input reaches it: log(x) reaches 100 at x = 10¹⁰⁰. That's why a log function has no horizontal asymptote.
Shifts and the input-ratio pattern
A horizontal shift, g(x) = log_b(x − h), moves the asymptote to x = h and the domain to x > h. A vertical shift, log_b(x) + k, doesn't change the domain or asymptote.
A pure log function has the property that equal ratios of inputs give equal changes in outputs. After a horizontal shift that property is lost for x itself, but it holds for x − h. So if (x − h) values change proportionally while the outputs change by equal amounts, the original function is logarithmic.
A vertical stretch, a · log_b(x), changes how steeply the graph rises or falls, but not its domain, its asymptote, or its x-intercept at (1, 0).
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Features of a basic log
Describe f(x) = log₂(x): domain, range, direction, concavity, intercept, asymptote and end behavior.
Show the solutionHide the solution
- Step 1: Points: (1/2, −1), (1, 0), (2, 1), (4, 2), (8, 3).
- Step 2: Domain x > 0; range all reals. The only x-intercept is (1, 0) because log₂(1) = 0.
- Step 3: Outputs rise as inputs rise: increasing. The rises get smaller for equal input steps (from 1 to 2 the output rises 1; from 2 to 3 it rises only about 0.585), so the rate of change decreases: concave down.
- Step 4: Vertical asymptote x = 0.
Answer: Domain x > 0, range all reals, increasing, concave down, x-intercept (1, 0), vertical asymptote x = 0; lim (x→0⁺) f(x) = −∞ and lim (x→∞) f(x) = ∞.
- Example 2
A shifted log
For g(x) = log₂(x − 3) + 1, find the domain, vertical asymptote and zero.
Show the solutionHide the solution
- Step 1: The input of the log must be positive: x − 3 > 0, so x > 3.
- Step 2: The asymptote is where the log's input is 0: x = 3.
- Step 3: Zero: log₂(x − 3) + 1 = 0, so log₂(x − 3) = −1. Rewrite: x − 3 = 2⁻¹ = 1/2, so x = 3.5.
Answer: Domain x > 3; vertical asymptote x = 3; zero at x = 7/2.
- Example 3
Trap: not every log rises
Describe h(x) = −2 log₃(x).
Show the solutionHide the solution
- Step 1: log₃(x) is increasing and concave down. Multiplying by −2 reflects it over the x-axis and stretches it.
- Step 2: So h is decreasing and concave up. Students who memorize “logs increase and are concave down” get this wrong.
- Step 3: End behavior flips too: as x → 0⁺, log₃(x) → −∞, so h(x) → ∞. As x → ∞, h(x) → −∞.
Answer: Decreasing and concave up, with vertical asymptote x = 0, lim (x→0⁺) h(x) = ∞ and lim (x→∞) h(x) = −∞.
Common mistakes
- Saying log functions have a horizontal asymptote because they grow slowly. They grow without bound; the only asymptote is vertical.
- Forgetting the domain after a shift. log(x + 5) needs x > −5, not x > 0.
- Assuming every log function is increasing and concave down. A negative a or a base between 0 and 1 flips both.
On the exam
- Expect questions on the domain, asymptote and end behavior of transformed log functions, often written with limit notation.
- Concavity questions may give a table and ask you to justify using how the rate of change behaves over equal intervals.
Connected topics
Videos
Check yourself
4 questions on 2.11 Logarithmic Functions. Pick an answer to see if you got it, and why.
Let f(x) = −2 log₃ x. Which of the following statements about f is true?
Let f(x) = ln(6 − 2x). Which of the following is true?
Let f(x) = log₅(x + 4) − 1. Which of the following gives the vertical asymptote and the zero of f?
| x | 1 | 3 | 9 | 27 |
|---|---|---|---|---|
| h(x) | 5 | 7 | 9 | 11 |
Table of values
Which of the following could define h?
0 of 4 answered