College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Section 4.1 - Exponential Functions - 4.1 Exercises - Page 372: 24



Work Step by Step

The graph of the function $f(x)=a^x$ contains the point $(-3,8)$. This means that when $x=-32$, $y=f(-3)=8$. Substitute $x$ and $y$ into $f(x) =a^x$ to obtain: $\begin{array}{ccc} \\&f(x) &= &a^x \\&f(-3) &= &a^{-3} \\&8 &= &a^{-3}\end{array}$ Using the rule $a^{-m} = \dfrac{1}{a^m}$, the expression above is equivalent to: $8=\dfrac{1}{a^3}$ Multiply $a^3$ to both sides of the equation to obtain: $8(a^3) = \dfrac{1}{a^3} \cdot a^3 \\8a^3=1$ Divide $8$ to both sides to obtain: $\dfrac{8a^3}{8} = \dfrac{1}{8} \\a^3=\dfrac{1}{8}$ Note that $\dfrac{1}{8} = \left(\dfrac{1}{2}\right)^3$. Thus, the expression above is equivalent to: $a^3 = \left(\dfrac{1}{2}\right)^3$ Use the rule "$a^m=b^m \longrightarrow a=b$" to obtain: $a=\dfrac{1}{2}$ With $a=\frac{1}{2}$, the function whose graph is given is $\color{blue}{f(x)=\left(\dfrac{1}{2}\right)^x}$.
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