Intermediate Algebra (12th Edition)

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

Chapter 7 - Review Exercises: 41

Answer

$s^{\frac{1}{2}}$

Work Step by Step

We are given that $(\sqrt[n] a)^{m}=\sqrt[n] a^{m}=a^{\frac{m}{n}}$, as long as all indicated roots are real numbers. Therefore, $\sqrt[8] s^{4}=s^{\frac{4}{8}}=s^{\frac{4\div4}{8\div4}}=s^{\frac{1}{2}}$.
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