Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 12 - Exponential Functions and Logarithmic Functions - 12.3 Logarithmic Functions - 12.3 Exercise Set - Page 803: 47

Answer

The graph is shown below.

Work Step by Step

$f\left( x \right)={{3}^{x}}$and ${{f}^{-1}}\left( x \right)={{\log }_{3}}x$ Substitute $x=0$ in$f\left( x \right)={{3}^{x}}$: $\begin{align} & f\left( 0 \right)={{3}^{0}} \\ & =1 \end{align}$ Substitute $x=1$ in$f\left( x \right)={{3}^{x}}$: $\begin{align} & f\left( 0 \right)={{3}^{1}} \\ & =3 \end{align}$ Substitute $x=2$ in$f\left( x \right)={{3}^{x}}$: $\begin{align} & f\left( 0 \right)={{3}^{2}} \\ & =9 \end{align}$ Substitute $x=-1$ in$f\left( x \right)={{3}^{x}}$: $\begin{align} & f\left( 0 \right)={{3}^{-1}} \\ & =\frac{1}{3} \end{align}$ Substitute $x=-2$ in$f\left( x \right)={{3}^{x}}$: $\begin{align} & f\left( 0 \right)={{3}^{-2}} \\ & =\frac{1}{9} \end{align}$ $\begin{matrix} x & f\left( x \right) \\ 0 & 1 \\ 1 & 3 \\ 2 & 9 \\ -1 & \frac{1}{3} \\ -2 & \frac{1}{9} \\ \end{matrix}$ Consider the second function, ${{f}^{-1}}\left( x \right)={{\log }_{3}}x$ Assume, ${{f}^{-1}}\left( x \right)=y$ The function $y={{\log }_{3}}x$ can be written as ${{3}^{y}}=x$. Substitute $y=0$ in $x={{3}^{y}}$: $\begin{align} & x={{3}^{0}} \\ & =1 \end{align}$ Substitute $y=1$ in $x={{3}^{y}}$: $\begin{align} & x={{3}^{1}} \\ & =3 \end{align}$ Substitute $y=2$ in $x={{3}^{y}}$: $\begin{align} & x={{3}^{2}} \\ & =9 \end{align}$ Substitute $y=-1$ in $x={{3}^{y}}$: $\begin{align} & x={{3}^{-1}} \\ & =\frac{1}{3} \end{align}$ Substitute $y=-2$ in $x={{3}^{y}}$: $\begin{align} & x={{3}^{-2}} \\ & =\frac{1}{9} \end{align}$ $\begin{matrix} {{f}^{-1}}\left( x \right) & y \\ 1 & 0 \\ 3 & 1 \\ 9 & 2 \\ \frac{1}{3} & -1 \\ \frac{1}{9} & -2 \\ \end{matrix}$ Now, draw the graph of the functions $f\left( x \right)={{3}^{x}}$ and ${{f}^{-1}}\left( x \right)={{\log }_{3}}x$.
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