Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.3 Exponential Functions - 5.3 Assess Your Understanding - Page 280: 1


$64, 4, \frac{1}{9}.$

Work Step by Step

$4^3=4\cdot4\cdot4=64$ Use the rule $a^{m/n}=(\sqrt[n]{a})^m$ to obtain: $8^{2/3}=(\sqrt[3] 8)^2=(\sqrt[3]{2^3})^2=2^24$ Use the rule $a^{-m} = \frac{1}{a^m}, m\gt 0$ to obtain: $3^{-2}=\frac{1}{3^2}=\frac{1}{9}$
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