College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Review Exercises - Page 499: 13

Answer

(a) $f(4)=81$ (b) $g(9)=2$ (c) $f(-2)=\frac{1}{9}$ (d) $g\left(\frac{1}{27}\right)=-3$

Work Step by Step

(a) Substitute $4$ to $x$ in $f(x)$ to obtain: \begin{align*} f(4)&=3^4\\ f(4)&=81 \end{align*} (b) Substitute $9$ to $x$ in $g(x)$ to obtain: \begin{align*} g(9)&=\log_3{9}\\ g(9)&=\log_3{(3^2)}\\ g(9)&=2 \end{align*} (c) Substitute $-2$ to $x$ in $f(x)$ to obtain: \begin{align*} f(-2)&=3^{-2}\\\\ f(-2)&=\frac{1}{3^2}\\\\ f(-2)&=\frac{1}{9}\\\\ \end{align*} (d) Substitute $\frac{1}{27}$ to $x$ in $g(x)$ to obtain: \begin{align*} g\left(\frac{1}{27}\right)&=\log_3{\left(\frac{1}{27}\right)}\\\\ &=\log_3{\left(\frac{1}{3^3}\right)}\\\\ &=\log_3{\left(3^{-3}\right)}\\\\ &=-3 \end{align*}
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