Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.3 - Logarithmic Functions - Exercise Set - Page 701: 76

Answer

$x=16$

Work Step by Step

RECALL: $\log_b{a}=y \longrightarrow b^y=a$ With $a=x$, $b=64$, and $y=\frac{2}{3}$, use the rule above to the given expression to obtain: $64^{\frac{2}{3}}=x$ Use the rule $a^{\frac{m}{n}} = (\sqrt[n]{a})^m$ to obtain: $(\sqrt[3]{64})^2=x \\(\sqrt[3]{4^3})^2=x$ Use the rule $\sqrt[3]{a^3} = a$ to obtain: $4^2=x \\4(4)=x \\16=x$
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