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 - Page 278: 137



Work Step by Step

Using the laws of exponents, the given expression, $ \left( \dfrac{5z^3}{2a^2} \right)^{-3} \left( \dfrac{8a^{-1}}{15z^{-2}} \right)^{-3} ,$ is equivalent to \begin{array}{l}\require{cancel} \left( \dfrac{2a^2}{5z^3} \right)^{3} \left( \dfrac{15z^{-2}}{8a^{-1}} \right)^{3} \\\\= \dfrac{2^3a^{2(3)}}{5^3z^{3(3)}}\cdot \dfrac{15^3z^{-2(3)}}{8^3a^{-1(3)}} \\\\= \dfrac{2^3a^{6}}{5^3z^{9}}\cdot \dfrac{15^3z^{-6}}{8^3a^{-3}} \\\\= \dfrac{2^3\cdot15^3a^{6}z^{-6}}{5^3\cdot8^3z^{9}a^{-3}} \\\\= \dfrac{2^3\cdot3^3\cdot\cancel{5^3}a^{6}z^{-6}}{\cancel{5^3}\cdot2^9z^{9}a^{-3}} \\\\= 2^{3-9}\cdot3^3\cdot a^{6-(-3)}z^{-6-9} \\\\= 2^{-6}\cdot27\cdot a^{9}z^{-15} \\\\= \dfrac{27a^{9}}{2^{6}z^{15}} \\\\= \dfrac{27a^{9}}{64z^{15}} .\end{array}
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