Intermediate Algebra (12th Edition)

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

Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises: 103

Answer

$\frac{32}{x^{5}}$

Work Step by Step

According to the definition of negative exponents, $a^{-n}=\frac{1}{a^{n}}$ (where $a\ne0$). Therefore, $(\frac{x}{2})^{-5}=\frac{1}{(\frac{x}{2})^{5}}=(\frac{2}{x})^{5}=\frac{2^{5}}{x^{5}}=\frac{32}{x^{5}}$.
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