Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Chapter 4 Test: 4

Answer

$\dfrac{16}{9p^{10}q^{28}}$

Work Step by Step

Use the laws of exponents to simplify the given expression, $ \left( \dfrac{4p^2}{q^4} \right)^3\left( \dfrac{6p^8}{q^{-8}} \right)^{-2} .$ Using the law of exponents which states that $\left( \dfrac{a^xb^y}{c^z} \right)^m=\dfrac{a^{xm}b^{ym}}{c^{zm}},$ the expression above simplifies to \begin{array}{l}\require{cancel} \left( \dfrac{4^3p^{2(3)}}{q^{4(3)}} \right)\left( \dfrac{6^{-2}p^{8(-2)}}{q^{-8(-2)}} \right) \\\\= \left( \dfrac{4^3p^{6}}{q^{12}} \right)\left( \dfrac{6^{-2}p^{-16}}{q^{16}} \right) \\\\= \dfrac{4^3p^{6}(6^{-2}p^{-16})}{q^{12}(q^{16})} .\end{array} Using the law of exponents which states that $a^x\cdot a^y=a^{x+y},$ the expression above simplifies to \begin{array}{l}\require{cancel} \dfrac{4^3(6^{-2})p^{6+(-16)}}{q^{12+16}} \\\\= \dfrac{4^3(6^{-2})p^{-10}}{q^{28}} .\end{array} Using the law of exponents which states that $a^{-x}=\dfrac{1}{a^x}$ or $\dfrac{1}{a^{-x}}=a^x$ the expression above simplifies to \begin{array}{l}\require{cancel} \dfrac{4^3}{6^{2}p^{10}q^{28}} \\\\= \dfrac{64}{36p^{10}q^{28}} \\\\= \dfrac{\cancel{4}(16)}{\cancel{4}(9)p^{10}q^{28}} \\\\= \dfrac{16}{9p^{10}q^{28}} .\end{array}
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