## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises - Page 278: 147

#### Answer

$-\dfrac{3}{32m^{8}p^{4}}$

#### Work Step by Step

Using the laws of exponents, the given expression, $\dfrac{(2m^2p^3)^2(4m^2p)^{-2}}{(-3mp^4)^{-1}(2m^3p^4)^3} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{(2^2m^{2(2)}p^{3(2)})(4^{-2}m^{2(-2)}p^{-2})}{(-3)^{-1}(m^{-1}p^{4(-1)})(2^3m^{3(3)}p^{4(3)})} \\\\= \dfrac{(4m^{4}p^{6})(4^{-2}m^{-4}p^{-2})}{(-3)^{-1}(m^{-1}p^{-4})(8m^{9}p^{12})} \\\\= \dfrac{4(-3)^{1}m^{4+(-4)}p^{6+(-2)}}{4^{2}(m^{-1+9}p^{-4+12})(8)} \\\\= \dfrac{4(-3)m^{0}p^{6-2}}{16(m^{8}p^{8})(8)} \\\\= \dfrac{-12m^{0}p^{4}}{128m^{8}p^{8}} \\\\= \dfrac{\cancel{4}(-3)m^{0-8}p^{4-8}}{\cancel{4}(32)} \\\\= \dfrac{-3m^{-8}p^{-4}}{32} \\\\= -\dfrac{3}{32m^{8}p^{4}} .\end{array}

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