College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Summary, Review, and Test - Review Exercises: 63

Answer

4

Work Step by Step

Based on the definition of $a^{\frac{m}{n}}$, we know that $a^{\frac{m}{n}}=(\sqrt[n] a)^{m}=\sqrt[n] a^{m}$ (where $\sqrt[n] a$ is a real number). Therefore, $16^{\frac{1}{2}}=\sqrt[2] 16^{1}=\sqrt 16=4$. We know that $\sqrt 16=4$, because $4^{2}=16$.
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