Precalculus (10th Edition)

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

Appendix A - Review - A.10 nth Roots; Rational Exponents - A.10 Assess Your Understanding - Page A88: 76

Answer

$\dfrac{9}{4} $

Work Step by Step

Use $27=3^3$ and $8=2^3$. $\left (\dfrac{27}{8} \right )^{\frac{2}{3}}=\left (\dfrac{3^3}{2^3} \right )^{\frac{2}{3}}$ Use the rule $\left(\dfrac{a}{b}\right)^n=\dfrac{a^m}{b^m}$: $=\dfrac{(3^3)^{\frac{2}{3}}}{(2^3)^{\frac{2}{3}}} $ Use the rule $\left(a^m\right)^n=a^{mn}$, then simplify to obtain: $=\dfrac{3^{3\cdot \frac{2}{3}}}{2^{3\cdot \frac{2}{3}}} $ $=\dfrac{3^{2}}{2^{2}} $ $=\dfrac{9}{4} $ Hence, the correct answer is $\frac{9}{4} $.
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