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: 64



Work Step by Step

We know that $b^{-n}=\frac{1}{b^{n}}$ (where $b$ is a nonzero real number and $n$ is a natural number). Therefore, $25^{-\frac{1}{2}}=\frac{1}{25^{\frac{1}{2}}}$ 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, $\frac{1}{25^{\frac{1}{2}}}=\frac{1}{\sqrt[2] 25^{1}}=\frac{1}{\sqrt 25}=\frac{1}{5}$ We know that $\sqrt 25=5$, because $5^{2}=25$.
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