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: 16

Answer

$\dfrac{27}{64}$

Work Step by Step

RECALL: (1) $a^{-m} = \dfrac{1}{a^m}$ (2) $\dfrac{a^m}{a^n} = a^{m-n}$ (3) $\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}$ Use rule (1) above to obtain: $\frac{4}{3})^{-3} = \dfrac{1}{(\frac{4}{3})^3}$ Use rule (3) above to obtain: $=\dfrac{1}{\frac{4^3}{3^3}} \\=\dfrac{1}{\frac{64}{27}}$ Use the rule $a \div \dfrac{b}{c} = a \cdot \dfrac{c}{b}$ to obtain: $=1 \cdot \dfrac{27}{64} \\=\dfrac{27}{64}$
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