College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.6 - Rational Exponents - R.6 Exercises: 35

Answer

$\dfrac{4}{a^2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $ \dfrac{4a^5(a^{-1})^3}{(a^{-2})^{-2}} .$ $\bf{\text{Solution Details:}}$ Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{4a^5a^{-1(3)}}{a^{-2(-2)}} \\\\= \dfrac{4a^5a^{-3}}{a^{4}} .\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{4a^{5+(-3)}}{a^{4}} \\\\= \dfrac{4a^{5-3}}{a^{4}} \\\\= \dfrac{4a^{2}}{a^{4}} .\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} 4a^{2-4} \\\\= 4a^{-2} .\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{4}{a^2} .\end{array}
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