Inverses of Exponential Functions

Part 1 · Hook

Why this matters

A savings balance doubles every

7

years. Asking "how much after

21

years?" uses an exponential function. Asking "when will it reach

8

times what I put in?" runs the same rule backward. That backward rule is a logarithm with the same base: exponentials and logarithms undo each other.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $\log_2 32$ ?

$5$
$16$
$6$
$\frac{1}{5}$

Show the answer

$2^5 = 32$ , so $\log_2 32 = 5$ .

Correct: $5$ :
$16$ :
$6$ :
$\frac{1}{5}$ :

2. The point $(3, 7)$ is on the graph of an invertible function $f$ . Which point is on the graph of $f^{-1}$ ?

$(7, 3)$
$(3, 7)$
$(-3, -7)$
$\left(3, \frac{1}{7}\right)$

Show the answer

An inverse swaps inputs and outputs.

Correct: $(7, 3)$ :
$(3, 7)$ :
$(-3, -7)$ :
$\left(3, \frac{1}{7}\right)$ :

Part 4 · See it

See it first

Graphs of y = 2 to the x and y = log base 2 of x, mirror images across the dashed line y = x, with (2, 4) matching (4, 2) and (0, 1) matching (1, 0). — $y = 2^x$ and $y = \log_2 x$ reflect across $y = x$ : $(2, 4)$ matches $(4, 2)$ , and the horizontal asymptote $y = 0$ matches the vertical asymptote $x = 0$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$f(x) = b^x$ sends the exponent $a$ to $b^a$ $g(x) = \log_b x$ sends $b^a$ back to $a$ : they are inverses
Compose them $\log_b(b^x) = x$ for every $x$ , and $b^{\log_b x} = x$ for $x > 0$
Swap coordinatesthe domain $(-\infty, \infty)$ and range $(0, \infty)$ of $b^x$ trade places, and $y = 0$ becomes $x = 0$
Add to the input of $b^x$ the output is multiplied
Multiply the input of $\log_b x$ the output increases by a fixed amount

Part 6 · Key ideas

Key ideas

$y = b^x$ and $y = \log_b x$ (same base) are inverse functions: $\log_b(b^x) = x$ and $b^{\log_b x} = x$ for $x > 0$ .
Their graphs reflect across $y = x$ ; $\ln x$ and $e^x$ are the most used pair.
$b^x$ : domain all real numbers, range $y > 0$ , asymptote $y = 0$ . $\log_b x$ : domain $x > 0$ , range all real numbers, asymptote $x = 0$ .
Exponential: equal additions to the input multiply the output. Logarithmic: equal factors on the input add to the output.

Part 7 · Misconception

A common mistake

The wrong idea: The inverse of $f(x) = 5^x$ is the fifth root, $\sqrt[5]{x}$ .

What actually happens: A fifth root undoes the power $x^5$ , where the variable is the base. In $5^x$ the variable is the exponent, so the inverse is $\log_5 x$ .

Part 8 · Check yourself

Check yourself

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

Table

Values of $h(x) = \log_4 x$

Values of h(x) = log base 4 of x at six inputs, each 4 times the one before
$x$	$\frac{1}{16}$	$\frac{1}{4}$	1	4	16	64
$h(x)$	-2	-1	0	1	2	3

1. In the table, each input is $4$ times the one before it. What happens to $h(x)$ at each step?

It increases by $1$
It is multiplied by $4$
It increases by $4$
It is multiplied by $\frac{1}{4}$

Show the answer

Multiplying the input of $\log_4$ by $4$ adds $1$ to the output: the outputs go $-2, -1, 0, 1, 2, 3$ .

Correct: It increases by $1$ : Equal factors in the input give equal differences in the output: $+1$ each time.
It is multiplied by $4$ : Multiplying by $4$ is what happens to the inputs, not the outputs.
It increases by $4$ : The outputs go up by $1$ , from $0$ to $1$ to $2$ , not by $4$ .
It is multiplied by $\frac{1}{4}$ : The outputs rise, from $0$ to $1$ to $2$ ; they are not divided by $4$ .

2. The inverse of $h$ is $k(x) = 4^x$ . Find $k(2.5)$ .

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

Show the answer

$k(2.5)$ $= 4^{5/2}$ $= (\sqrt{4})^5$ $= 2^5$ $= 32$ . It lies between $k(2) = 16$ and $k(3) = 64$ , but not halfway: equal input steps multiply outputs.

Answer: 32

3. The inverse of $h$ is $k(x) = 4^x$ . Which range does $k$ have?

$(0, \infty)$
$(-\infty, \infty)$
$[1, \infty)$
$(-\infty, 0)$

Show the answer

The range of the inverse is the domain of $h$ . A base- $4$ logarithm takes only positive inputs, so $k$ has only positive outputs.

Correct: $(0, \infty)$ : Domain of $h$ , $x > 0$ , becomes the range of $k$ .
$(-\infty, \infty)$ : That is the range of $h$ , which becomes the domain of $k$ .
$[1, \infty)$ : $k(-1) = \frac{1}{4}$ , an output less than $1$ .
$(-\infty, 0)$ : Powers of $4$ are positive, never negative.

Graph

The graph of $f(x) = 3^x$

g

is the inverse of

f(x) = 3^x

, so

g(x) = \log_3 x

.

4. The point $(1, 3)$ is on the graph of $f$ . Which point is on the graph of $g$ ?

$(3, 1)$
$(1, 3)$
$\left(1, \frac{1}{3}\right)$
$(-1, -3)$

Show the answer

$g$ swaps inputs and outputs of $f$ : $f(1) = 3$ , so $g(3) = 1$ .

Correct: $(3, 1)$ : $\log_3 3 = 1$ .
$(1, 3)$ : This point is on $f$ itself.
$\left(1, \frac{1}{3}\right)$ : $\frac{1}{3}$ is a reciprocal, not a swap of coordinates.
$(-1, -3)$ : That reflects across the origin, not across $y = x$ .

5. The graph of $f$ has the horizontal asymptote $y = 0$ . Which statement describes $g$ near its asymptote?

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

Show the answer

Reflecting across $y = x$ turns the horizontal asymptote $y = 0$ into the vertical asymptote $x = 0$ . As $x$ approaches $0$ from the right, $\log_3 x$ decreases without bound.

Correct: $\lim_{x \to 0^+} g(x) = -\infty$ : Small positive inputs, such as $\frac{1}{9}$ , give large negative outputs, such as $-2$ .
$\lim_{x \to -\infty} g(x) = 0$ : That describes $f$ : $\lim_{x \to -\infty} 3^x = 0$ .
$\lim_{x \to 0^+} g(x) = 0$ : $g(1) = 0$ ; near $x = 0$ the outputs fall without bound.
$\lim_{x \to \infty} g(x) = 0$ : $g$ keeps growing, slowly: $g(81) = 4$ .

6. Write the inverse of $f(x) = 5^x$ .

Type an expression for $f^{-1}(x) =$ .

Show the answer

The base- $5$ logarithm undoes the base- $5$ exponential: $f^{-1}(x) = \log_5 x$ .

Answer: $f^{-1}(x) =$ $\log_{5}(x)$

7. Let $f(x) = 2^x$ . When $x$ increases by $3$ , what happens to $f(x)$ ?

It is multiplied by $8$
It increases by $8$
It is multiplied by $6$
It increases by $3$

Show the answer

$f(x + 3)$ $= 2^{x + 3}$ $= 2^x \cdot 2^3$ $= 8 f(x)$ . Adding to the input multiplies the output.

Correct: It is multiplied by $8$ : $2^{x + 3} = 8 \cdot 2^x$ .
It increases by $8$ : From $x = 0$ to $x = 3$ the output goes from $1$ to $8$ , up $7$ ; the change grows with $x$ . The factor is always $8$ .
It is multiplied by $6$ : $2 \cdot 3 = 6$ , but three factors of $2$ make $2^3 = 8$ .
It increases by $3$ : That is the change in the input, not in the output.

Part 9 · Summary

Summary

With the same base,

f(x) = b^x

and

g(x) = \log_b x

are inverses: each undoes the other, their graphs reflect across

y = x

, and their domains, ranges and asymptotes trade places. Adding to the input of an exponential multiplies its output; multiplying the input of a logarithm adds to its output. To invert a stretched exponential, undo the outer steps first, then take the logarithm.

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

Values of h(x)=log4⁡xh(x) = \log_4 x

The graph of f(x)=3xf(x) = 3^x

Summary

What comes next

Connections

Values of $h(x) = \log_4 x$

The graph of $f(x) = 3^x$