Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 5 - Section 5.7 - Negative Exponents and Scientific Notation - Exercise Set - Page 408: 18

Answer

$\dfrac{1}{16}$

Work Step by Step

Using $x^{-a}=\dfrac{1}{x^a}$ and $\dfrac{1}{x^{-a}}=x^a,$ the given expression, $ \dfrac{4^{-3}}{2^{-2}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{2^{2}}{4^{3}} \\\\= \dfrac{4}{64} \\\\= \dfrac{\cancel{4}}{\cancel{4}\cdot16} \\\\= \dfrac{1}{16} .\end{array}
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