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: 45

Answer

The graph is shown below.

Work Step by Step

$f\left( x \right)={{\log }_{2.5}}x$ Assume, $f\left( x \right)=y$ Therefore, the function becomes: $y={{\log }_{2.5}}x$ Therefore, the function $y={{\log }_{2.5}}x$ can be written as $x={{2.5}^{y}}$ Substitute $y=0,1,2,-1,-2$ in the function ${{2.5}^{y}}=x$ and tabulate the values as shown in Table 1. Substitute in a selected y value: $\begin{align} & x={{\left( 2.5 \right)}^{0}} \\ & =1 \end{align}$ Substitute in a selected y value: $\begin{align} & x={{\left( 2.5 \right)}^{1}} \\ & =2.5 \end{align}$ Substitute in a selected y value: $\begin{align} & x={{\left( 2.5 \right)}^{2}} \\ & =6.25 \end{align}$ Substitute in a selected y value: $\begin{align} & x={{\left( 2.5 \right)}^{-1}} \\ & =\frac{1}{2.5} \end{align}$ Substitute in a selected y value: $\begin{align} & x={{\left( 2.5 \right)}^{-2}} \\ & =\frac{1}{{{2.5}^{2}}} \\ & =\frac{1}{6.25} \end{align}$ Tabulate the obtained values as shown below: $\begin{matrix} {{2.5}^{y}}=x & y \\ 1 & 0 \\ 2.5 & 1 \\ 6.25 & 2 \\ 0.4 & -1 \\ 0.16 & -2 \\ \end{matrix}$
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