Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.6 Rational Exponents - R.6 Exercises - Page 63: 61

Answer

$\color{blue}{\dfrac{256}{81}}$

Work Step by Step

RECALL: (1) $a^{m/n} = \left(a^{1/n}\right)^m$ (2) $a^{1/n} = \sqrt[n]{a}$ (3) For positive real numbers $a$, $\sqrt[n]{a^n}=a$ (4) $a^{-m} = \dfrac{1}{a^m}$ (5) When $n$ is odd, $\sqrt[n]{a^n} = n$ (6) $\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}$ Use rule (4) above to obtain: $\left(\dfrac{27}{64}\right)^{-4/3} = \dfrac{1}{(\frac{27}{64})^{4/3}}$ Use rule (1) above to obtain: $=\dfrac{1}{[(\frac{27}{64})^{1/3}]^4}$ Use rule (2) above to obtain: $=\dfrac{1}{\left(\sqrt[3]{\frac{27}{64}}\right)^4}$ With $\dfrac{27}{64}=\left(\dfrac{3}{4}\right)^3$, the expression above is equivalent to: $=\dfrac{1}{\left(\sqrt[3]{(\frac{3}{4})^3}\right)^4}$ Use rule (5) above to obtain: $=\dfrac{1}{\left(\frac{3}{4}\right)^4}$ Use rule (6) above to obtain: $\\=\dfrac{1}{\frac{3^4}{4^4}} \\=\dfrac{1}{\frac{81}{256}} \\=1 \cdot \dfrac{256}{81} \\=\color{blue}{\dfrac{256}{81}}$
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