2.10 - Inverses of Exponential Functions

Overview

Exponential functions are one-to-one, so they have inverses, and those inverses are the logarithmic functions. This topic shows how

b^x

and

\log_b x

undo each other, why their tables and graphs are mirror images, and how their patterns of change are reversed: exponential outputs multiply when inputs add, and logarithmic outputs add when inputs multiply.

The general form of a logarithmic function

A logarithmic function in general form is

f(x) = a \log_b x, \quad \text{where } b > 0,\ b \neq 1, \text{ and } a \neq 0

$b$ is the base, with the same rules as in Topic 2.9.

$a$ is a vertical dilation. It can't be 0, or the function would just be the constant 0. If $a < 0$ , the graph is also reflected over the $x$ -axis.

When

a = 1

, you have the basic logarithmic function

f(x) = \log_b x

Logarithms and exponentials are inverses

An exponential function with an initial value of 1,

g(x) = b^x

, is always increasing (when

b > 1

) or always decreasing (when

0 < b < 1

). So it's one-to-one and has an inverse. That inverse is the logarithm with the same base:

g(x) = b^x \quad \text{and} \quad f(x) = \log_b x \quad \text{are inverse functions}

This comes straight from the definition in Topic 2.9. Using the swap method from Topic 2.8:

y = b^x \;\xrightarrow{\text{swap}}\; x = b^y \;\Longleftrightarrow\; y = \log_b x

Since they're inverses, composing them in either order gives the identity function:

\log_b\!\left(b^x\right) = x \text{ for all real } x \qquad b^{\log_b x} = x \text{ for } x > 0

For example,

\log_3(3^7) = 7

10^{\log 45} = 45

, and

2^{\log_2 13} = 13

Ordered pairs flip

(s, t)

is an ordered pair of

g(x) = b^x

, then

(t, s)

is an ordered pair of

f(x) = \log_b x

. Here's base 2:

$x$	$-2$	$-1$	0	1	2	3
$g(x) = 2^x$	$\frac{1}{4}$	$\frac{1}{2}$	1	2	4	8

$x$	$\frac{1}{4}$	$\frac{1}{2}$	1	2	4	8
$f(x) = \log_2 x$	$-2$	$-1$	0	1	2	3

The pair

(3, 8)

g

matches the pair

(8, 3)

f

, which says

\log_2 8 = 3

. Two pairs come free for every base:

(0, 1)

b^x

gives

(1, 0)

\log_b x

, and

(1, b)

b^x

gives

(b, 1)

\log_b x

Graphs: A reflection over $y = x$

Because they're inverses, the graph of

f(x) = \log_b x

is the reflection of the graph of

g(x) = b^x

over the line

y = x

, the graph of the identity function

h(x) = x

2.10 - Inverses of Exponential Functions

Every feature flips along with the points. The domain and range trade places:

2^x

accepts every real number and gives only positive outputs, while

\log_2 x

accepts only positive inputs and gives every real output. The key points

(0, 1)

and

(1, 2)

become

(1, 0)

and

(2, 1)

, and the horizontal asymptote

y = 0

of the exponential becomes the vertical asymptote

x = 0

of the logarithm. Topic 2.11 lists these features for every base.

How inputs and outputs change together

Exponential and logarithmic functions have reversed patterns of change.

Exponential, $g(x) = ab^x$ : as the inputs increase in equal-length steps (add the same amount), the outputs change proportionally (multiply by the same factor). Inputs add, outputs multiply.

Logarithmic, $f(x) = a\log_b x$ : as the outputs increase in equal-length steps, the inputs change proportionally. Said the other way, when the inputs multiply by the same factor, the outputs add the same amount. Inputs multiply, outputs add.

For

f(x) = \log_2 x

, each time

x

doubles (1, 2, 4, 8, 16), the output goes up by 1 (0, 1, 2, 3, 4). For

f(x) = 3\log_2 x

, each doubling of

x

adds 3 to the output.

This is why logarithms grow so slowly. To raise

\log_2 x

from 10 to 20,

x

has to go from

2^{10} = 1{,}024

all the way to

2^{20} = 1{,}048{,}576

To spot a logarithmic pattern in a table, check whether the inputs have a constant ratio while the outputs change by a constant amount.

Example 1: The inverse of an exponential function

Let

g(x) = 5^x

. (a) Write the inverse of

g

. (b) The point

(2, 25)

is on the graph of

g

. What point is on the graph of

g^{-1}

? (c) Find

g^{-1}\left(\frac{1}{5}\right)

and

g^{-1}(\sqrt{5})

Part (a).

g^{-1}(x) = \log_5 x

Part (b). Flip the pair:

(25, 2)

. This says

\log_5 25 = 2

Part (c). Since

5^{-1} = \frac{1}{5}

g^{-1}\left(\frac{1}{5}\right) = \log_5 \frac{1}{5} = -1

. Since

5^{1/2} = \sqrt{5}

g^{-1}(\sqrt{5}) = \log_5 \sqrt{5} = \frac{1}{2}

Example 2: Finding the base from a point

The graph of

g(x) = b^x

passes through

(3, 64)

. Write the inverse of

g

and name the point on its graph that matches

(3, 64)

Since

g(3) = 64

, we need

b^3 = 64

. Because

4^3 = 64

, the base is

b = 4

. So

g(x) = 4^x

and

g^{-1}(x) = \log_4 x

The point

(3, 64)

g

becomes

(64, 3)

g^{-1}

, which checks out:

\log_4 64 = 3

Example 3: Recognizing a logarithmic pattern (AP-style)

Values of a function

h

are shown in the table.

$x$	1	3	9	27	81
$h(x)$	0	2	4	6	8

(a) Explain why a logarithmic function is a good model for

h

. (b) Write a formula for

h

in the form

h(x) = a\log_3 x

. (c) Find

h\left(\frac{1}{3}\right)

and the input

x

with

h(x) = 10

Part (a). As the outputs increase in equal steps of 2, the inputs are multiplied by 3 each time (1, 3, 9, 27, 81). The inputs change proportionally over equal-length output intervals, which is the pattern of a logarithmic function.

Part (b). Use the point

(3, 2)

h(3) = a\log_3 3 = a(1) = a = 2

h(x) = 2\log_3 x

. Check another point:

h(27) = 2\log_3 27 = 2(3) = 6

. It matches.