Graph f(x)=7^x , f^-1(x) = log base 7 of x

$f (x) = 7^{x}$ , $f^{- 1} (x) = \log_{7} (x)$

Step 1

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

Horizontal Asymptote: $y = 0$

Step 2

Graph $f^{- 1} (x) = \log_{7} (x)$ .

Find the asymptotes.

Set the argument of the logarithm equal to zero.

$x = 0$

The vertical asymptote occurs at $x = 0$ .

Vertical Asymptote: $x = 0$

Find the point at $x = 1$ .

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

$f (1) = \log_{7} (1)$

Simplify the result.

Logarithm base $7$ of $1$ is $0$ .

$f (1) = 0$

The final answer is $0$ .

$0$

Convert $0$ to decimal.

$y = 0$

Find the point at $x = 7$ .

Replace the variable $x$ with $7$ in the expression.

$f (7) = \log_{7} (7)$

Simplify the result.

Logarithm base $7$ of $7$ is $1$ .

$f (7) = 1$

The final answer is $1$ .

$1$

Convert $1$ to decimal.

$y = 1$

Find the point at $x = 2$ .

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

$f (2) = \log_{7} (2)$

The final answer is $\log_{7} (2)$ .

$\log_{7} (2)$

Convert $\log_{7} (2)$ to decimal.

$y = 0.35620718$

The log function can be graphed using the vertical asymptote at $x = 0$ and the points $(1, 0), (7, 1), (2, 0.35620718)$ .

Vertical Asymptote: $x = 0$

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 0.356 \\ 7 & 1 \end{array}$

Vertical Asymptote: $x = 0$

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 0.356 \\ 7 & 1 \end{array}$

Step 3

Plot each graph on the same coordinate system.

$f (x) = 7^{x}$

$f^{- 1} (x) = \log_{7} (x)$

Step 4

Graph f(x)=7^x , f^-1(x) = log base 7 of x

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