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 409: 58



Work Step by Step

Using the laws of exponents, the given expression, $ \left( 4x^{-1} \right)^{-2} ,$ simplifies to \begin{array}{l}\require{cancel} 4^{1(-2)}x^{-1(-2)} \\\\= 4^{-2}x^{2} \\\\= \dfrac{x^{2}}{4^{2}} \\\\= \dfrac{x^{2}}{16} .\end{array}
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