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 - Page 56: 33

Answer

$\dfrac{1}{2pq}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $ \dfrac{(3pq)q^2}{6p^2q^4} .$ $\bf{\text{Solution Details:}}$ 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{3pq^{1+2}}{6p^2q^4} \\\\= \dfrac{\cancel3pq^{3}}{\cancel3(2)p^2q^4} \\\\= \dfrac{pq^{3}}{2p^2q^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} \dfrac{p^{1-2}q^{3-4}}{2} \\\\= \dfrac{p^{-1}q^{-1}}{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{1}{2p^{1}q^{1}} \\\\= \dfrac{1}{2pq} .\end{array}
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