## College Algebra (11th Edition)

Published by Pearson

# Chapter R - Section R.6 - Rational Exponents - R.6 Exercises: 38

#### Answer

$\dfrac{1}{y^{10}}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $\dfrac{(8y^2)^{-4}(8y^5)^{-2}}{(8^{-3}y^{-4})^2} .$ $\bf{\text{Solution Details:}}$ Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{8^{-4}y^{2(-4)}8^{-2}y^{5(-2)}}{8^{-3(2)}y^{-4(2)}} \\\\= \dfrac{8^{-4}y^{-8}8^{-2}y^{-10}}{8^{-6}y^{-8}} .\end{array} Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{8^{-4+(-2)}y^{-8+(-10)}}{8^{-6}y^{-8}} \\\\= \dfrac{8^{-4-2}y^{-8-10}}{8^{-6}y^{-8}} \\\\= \dfrac{8^{-6}y^{-18}}{8^{-6}y^{-8}} \\\\= \dfrac{\cancel{8^{-6}}y^{-18}}{\cancel{8^{-6}}y^{-8}} \\\\= \dfrac{y^{-18}}{y^{-8}} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} y^{-18-(-8)} \\\\= y^{-18+8} \\\\= y^{-10} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{y^{10}} .\end{array}

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