Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.3 - Radicals and Rational Exponents - Exercise Set: 89

Answer

$\dfrac{1}{16}$

Work Step by Step

RECALL: (i) $a^{\frac{m}{n}}=(\sqrt[n]{a})^m.$ (ii) $a^{-m} = \dfrac{1}{a^m}, a\ne0.$ Use rule (ii) above to obtain $\dfrac{1}{32^{\frac{4}{5}}}.$ Use rule (i) with m=4 and n=5 to obtain $\dfrac{1}{(\sqrt[5]{32})^4} \\=\dfrac{1}{(\sqrt[5]{2^5})^4} \\=\dfrac{1}{(2)^4} \\=\dfrac{1}{16}.$
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