Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Review Exercises: 25

Answer

32

Work Step by Step

We are given that $a^{\frac{m}{n}}=\sqrt[n] a^{m}=(\sqrt[n] a)^{m}$, if all indicated roots are real numbers. Therefore, $16^{\frac{5}{4}}=\sqrt[4] 16^{5}=(\sqrt[4] 16)^{5}=(2)^{5}=32$. We know that $\sqrt[4] 16=2$, because $2^{4}=16$.
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