Inverses of Exponential Functions

Unit 2 · Exponential and Logarithmic Functions

What AP Precalc asks here

The exponential function $y = b^x$ and the logarithmic function $y = \log_b x$ are inverses. Composing them gives the identity: $b^{\log_b x} = x$ for $x > 0$ , and $\log_b(b^x) = x$ for all real $x$ . Graphically, $\log_b x$ is the reflection of $b^x$ across $y = x$ . The exponential's horizontal asymptote (HA) $y = 0$ becomes the log's vertical asymptote (VA) $x = 0$ , and domain/range swap.

Inverse identities

Exp(log)

b^{\log_b x} = x \quad (x > 0)

Log(exp)

\log_b(b^x) = x \quad (\text{all real } x)

Graph correspondence

Reflection

y = \log_b x \text{ is } y = b^x \text{ reflected across } y = x

Asymptotes

b^x: \; \text{HA } y = 0 \quad\longleftrightarrow\quad \log_b x: \; \text{VA } x = 0

AP Tip: Point swap is the fastest way to convert between exponential and log graphs. If

(2, 9)

is on

y = 3^x

, then

(9, 2)

is on

y = \log_3 x

Type 1

Simplify using inverse composition

Apply the identity directly. $b^{\log_b x} = x$ ; $\log_b(b^x) = x$ .

Example 1

Simplify

5^{\log_5 12}

. (A)

5

(B)

12

(C)

\log_5 12

(D)

60

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Type 2

Graph correspondence

Swap the coordinates of a point on the exponential to get a point on the log.

Example 2

(2, 9)

is on the graph of

y = 3^x

, what point is on the graph of

y = \log_3 x

? (A)

(-2, -9)

(B)

(2, 9)

(C)

(9, 2)

(D)

(1/2, 1/9)

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Type 3

Asymptote and domain/range correspondence

Exponentials have a horizontal asymptote at $y = 0$ ; their logs have a vertical asymptote at $x = 0$ . Domain and range swap.

Example 3

State the asymptotes of

y = e^x

and

y = \ln x

. (A)

e^x

: HA

y = 0

;

\ln x

: VA

x = 0

(B)

e^x

: VA

x = 0

;

\ln x

: HA

y = 0

y = 1

(D) Neither has an asymptote

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