Logarithmic Functions — AP® Precalculus Lesson

Part 1 · Hook

Why this matters

Suppose each time a phone's storage doubles, the price climbs by about the same amount. Doubling the input adds a fixed step to the output: that is how a logarithmic function grows. It rises without limit, but more and more slowly, and it has a wall on its left where the inputs run out.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $\log_2 8$ ?

$3$
$4$
$16$
$6$

Show the answer

$2^3 = 8$ .

Correct: $3$ :
$4$ :
$16$ :
$6$ :

2. $y = 2^x$ has the horizontal asymptote $y = 0$ . Which asymptote does its inverse $y = \log_2 x$ have?

$x = 0$
$y = 0$
$x = 1$
$y = 1$

Show the answer

Reflecting across $y = x$ turns $y = 0$ into $x = 0$ .

Correct: $x = 0$ :
$y = 0$ :
$x = 1$ :
$y = 1$ :

3. $\lim_{x \to 3^+} f(x) = -\infty$ says what about the graph of $f$ ?

Just right of $x = 3$ , it drops without bound
Just left of $x = 3$ , it drops without bound
It approaches $y = 3$ from above
It crosses $x = 3$ going down

Show the answer

$3^+$ means from the right; $-\infty$ means decreasing without bound.

Correct: Just right of $x = 3$ , it drops without bound:
Just left of $x = 3$ , it drops without bound:
It approaches $y = 3$ from above:
It crosses $x = 3$ going down:

Part 4 · See it

See it first

Graph of g(x) = 2 log base 2 of (x - 1), with a dashed vertical asymptote at x = 1 and points (2, 0), (3, 2), (5, 4) and (9, 6). — $g(x) = 2\log_2(x - 1)$ : the domain is $x > 1$ , $\lim_{x \to 1^+} g(x) = -\infty$ , and each doubling of $x - 1$ adds $2$ to the output. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The input of a logarithm must be positive $f(x)$ $= a\log_b(x - h) + k$ has domain $x > h$ and a vertical asymptote $x = h$
Inputs approach the asymptote from the rightfor $a > 0$ and $b > 1$ , $\lim_{x \to h^+} f(x) = -\infty$
Inputs grow without boundthe outputs grow without bound too, slowly: $\lim_{x \to \infty} f(x) = \infty$
Compare average rates over equal intervalsthey decrease, so $f$ is concave down; it is increasing with no maximum or minimum

Part 6 · Key ideas

Key ideas

$f(x) = a\log_b x$ ( $a \ne 0$ , $b > 0$ , $b \ne 1$ ): domain $x > 0$ , range all real numbers, vertical asymptote $x = 0$ .
For $a > 0$ , $b > 1$ : increasing, concave down, $\lim_{x \to 0^+} f(x) = -\infty$ and $\lim_{x \to \infty} f(x) = \infty$ . No extrema, no points of inflection.
Shifts move the asymptote: $\log_b(x - h)$ has domain $x > h$ and asymptote $x = h$ .
Write behavior near the asymptote with a one-sided limit from the side where $f$ is defined.

Part 7 · Misconception

A common mistake

The wrong idea: Because $\ln x$ grows so slowly, it levels off at a horizontal asymptote.

What actually happens: It never levels off: $\ln x$ reaches any number $n$ at $x = e^n$ , so $\lim_{x \to \infty} \ln x = \infty$ . Its rate of change decreases toward $0$ , but the outputs have no ceiling.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of $h$

The graph shows

h(x) = \log_2(x + 4) - 1

and its vertical asymptote

x = -4

.

1. What is the domain of $h$ ?

$x > -4$
$x > 0$
$x \ge -4$
$x > 4$

Show the answer

A logarithm needs a positive input: $x + 4 > 0$ , so $x > -4$ . The asymptote $x = -4$ itself is not included.

Correct: $x > -4$ : $x + 4 > 0$ gives $x > -4$ .
$x > 0$ : That is the domain of $\log_2 x$ ; the graph is shifted $4$ units left.
$x \ge -4$ : At $x = -4$ the input of the logarithm is $0$ , which is not allowed.
$x > 4$ : The shift is to the left: $x + 4 > 0$ .

2. Which statement describes $h$ near its vertical asymptote?

$\lim_{x \to -4^+} h(x)$ $= -\infty$
$\lim_{x \to -4^-} h(x)$ $= -\infty$
$\lim_{x \to -4^+} h(x) = \infty$
$\lim_{x \to -\infty} h(x) = -4$

Show the answer

$h$ is defined only to the right of $x = -4$ , and there the graph drops without bound.

Correct: $\lim_{x \to -4^+} h(x)$ $= -\infty$ : From the right, the outputs decrease without bound.
$\lim_{x \to -4^-} h(x)$ $= -\infty$ : $h$ is undefined to the left of $-4$ , so only the right-hand limit makes sense.
$\lim_{x \to -4^+} h(x) = \infty$ : The graph goes down, not up, as $x$ approaches $-4$ from the right.
$\lim_{x \to -\infty} h(x) = -4$ : $h$ is undefined for very negative $x$ ; $-4$ is where the asymptote is, not a limit value.

3. Find the zero of $h$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\log_2(x + 4) - 1 = 0$ means $\log_2(x + 4) = 1$ , so $x + 4 = 2$ and $x = -2$ . The graph crosses the $x$ -axis at $(-2, 0)$ .

Answer: -2

4. Which statement describes $h$ on its whole domain?

$h$ is increasing, and the rate of change of $h$ is decreasing
$h$ is increasing, and the rate of change of $h$ is increasing
$h$ is decreasing, and the rate of change of $h$ is increasing
$h$ has a maximum value at $x = 4$

Show the answer

The graph rises everywhere and bends downward: secant slopes get smaller, from $1$ on $[-3, -2]$ to $\frac{1}{2}$ on $[-2, 0]$ to $\frac{1}{4}$ on $[0, 4]$ . $h$ is concave down.

Correct: $h$ is increasing, and the rate of change of $h$ is decreasing: Rising outputs with shrinking secant slopes: increasing and concave down.
$h$ is increasing, and the rate of change of $h$ is increasing: An increasing rate of change would make the graph bend upward.
$h$ is decreasing, and the rate of change of $h$ is increasing: Larger inputs give larger outputs, so $h$ is increasing.
$h$ has a maximum value at $x = 4$ : A logarithmic function has no maximum; $h$ keeps growing past $x = 4$ .

5. Which statement describes the end behavior of $f(x) = \log_2 x$ ?

$\lim_{x \to \infty} \log_2 x$ $= \infty$
$\lim_{x \to \infty} \log_2 x$ $= 0$
$\lim_{x \to \infty} \log_2 x$ $= 2$
$\lim_{x \to \infty} \log_2 x$ $= 1$

Show the answer

Slowly but without bound: $\log_2 1{,}024 = 10$ and $\log_2 2^{100} = 100$ .

Correct: $\lim_{x \to \infty} \log_2 x$ $= \infty$ : Every whole number $n$ is reached at $x = 2^n$ .
$\lim_{x \to \infty} \log_2 x$ $= 0$ : $0$ is the value at $x = 1$ ; the outputs keep growing.
$\lim_{x \to \infty} \log_2 x$ $= 2$ : $\log_2 4 = 2$ and $\log_2 8 = 3$ : the outputs pass $2$ .
$\lim_{x \to \infty} \log_2 x$ $= 1$ : $\log_2 2 = 1$ , and the outputs pass $1$ .

6. The graph of $g$ is the graph of $y = \ln x$ shifted $3$ units right and $2$ units up. Write $g(x)$ .

Type an expression for $g(x) =$ .

Show the answer

A shift $3$ units right replaces $x$ by $x - 3$ ; a shift up adds $2$ to the output: $g(x) = \ln(x - 3) + 2$ .

Answer: $g(x) =$ $\ln(x - 3) + 2$

7. What is the domain of $f(x) = \log_3(x^2 - 9)$ ?

$x < -3$ or $x > 3$
$x > 3$
$x > 9$
$-3 < x < 3$

Show the answer

The input must be positive: $x^2 - 9 > 0$ , so $x^2 > 9$ , which holds for $x > 3$ and for $x < -3$ .

Correct: $x < -3$ or $x > 3$ : $x^2 > 9$ on both sides: $x > 3$ or $x < -3$ .
$x > 3$ : $x = -4$ also works: $(-4)^2 - 9 = 7 > 0$ .
$x > 9$ : The condition is $x^2 > 9$ , not $x > 9$ .
$-3 < x < 3$ : For $-3 < x < 3$ , $x^2 - 9$ is negative, so the logarithm is undefined there.

Part 9 · Summary

Summary

A logarithmic function

f(x) = a\log_b x

is defined only for

x > 0

, with a vertical asymptote at

x = 0

. With

a > 0

and

b > 1

it is increasing and concave down, so its rate of change decreases:

\lim_{x \to 0^+} f(x) = -\infty

and

\lim_{x \to \infty} f(x) = \infty

, with no extrema. Shifts move the asymptote and the domain. Describe asymptote behavior in limit notation from the side where

f

is defined.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of hh

Summary

What comes next

Connections

The graph of $h$