Algebra Examples

$f (x) = 9^{x}$ , $f^{- 1} (x) = \log_{9} (x)$

Step 1

Exponential functions have a horizontal asymptote. The equation of the horizontal asymptote is $y = 0$ .

Horizontal Asymptote: $y = 0$

Step 2

Graph $f^{- 1} (x) = \log_{9} (x)$ .

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Step 2.1

Find the asymptotes.

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Step 2.1.1

Set the argument of the logarithm equal to zero.

$x = 0$

Step 2.1.2

The vertical asymptote occurs at $x = 0$ .

Vertical Asymptote: $x = 0$

Step 2.2

Find the point at $x = 1$ .

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Step 2.2.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = \log_{9} (1)$

Step 2.2.2

Simplify the result.

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Step 2.2.2.1

Logarithm base $9$ of $1$ is $0$ .

$f (1) = 0$

Step 2.2.2.2

The final answer is $0$ .

$0$

Step 2.2.3

Convert $0$ to decimal.

$y = 0$

Step 2.3

Find the point at $x = 9$ .

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Step 2.3.1

Replace the variable $x$ with $9$ in the expression.

$f (9) = \log_{9} (9)$

Step 2.3.2

Simplify the result.

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Step 2.3.2.1

Logarithm base $9$ of $9$ is $1$ .

$f (9) = 1$

Step 2.3.2.2

The final answer is $1$ .

$1$

Step 2.3.3

Convert $1$ to decimal.

$y = 1$

Step 2.4

Find the point at $x = 2$ .

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Step 2.4.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = \log_{9} (2)$

Step 2.4.2

The final answer is $\log_{9} (2)$ .

$\log_{9} (2)$

Step 2.4.3

Convert $\log_{9} (2)$ to decimal.

$y = 0.31546487$

Step 2.5

The log function can be graphed using the vertical asymptote at $x = 0$ and the points $(1, 0), (9, 1), (2, 0.31546487)$ .

Vertical Asymptote: $x = 0$

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 0.315 \\ 9 & 1 \end{array}$

Vertical Asymptote: $x = 0$

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 0.315 \\ 9 & 1 \end{array}$

Step 3

Plot each graph on the same coordinate system.

$f (x) = 9^{x}$

$f^{- 1} (x) = \log_{9} (x)$

Step 4