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: 55

Answer

$\dfrac{1}{16x^{12}}$

Work Step by Step

Using the laws of exponents, the given expression, $ \left( \dfrac{4x^5}{2x^2} \right)^{-4} ,$ simplifies to \begin{array}{l}\require{cancel} \left( (4\div2)x^{5-2} \right)^{-4} \\\\= \left( 2x^{3} \right)^{-4} \\\\= 2^{1(-4)}x^{3(-4)} \\\\= 2^{-4}x^{-12} \\\\= \dfrac{1}{2^{4}x^{12}} \\\\= \dfrac{1}{16x^{12}} .\end{array}
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