College Algebra (11th Edition)

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

Chapter R - Review Exercises: 101

Answer

$\dfrac{1}{p^2q^4}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $ \dfrac{(p^{15}q^{12})^{-4/3}}{(p^{24}q^{16})^{-3/4}} .$ $\bf{\text{Solution Details:}}$ 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{p^{15\left(-\frac{4}{3} \right)}q^{12\left(-\frac{4}{3} \right)}}{p^{24\left(-\frac{3}{4} \right)}q^{16\left(-\frac{3}{4} \right)}} \\\\= \dfrac{p^{-20}q^{-16}}{p^{-18}q^{-12}} .\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} p^{-20-(-18)}q^{-16-(-12)} \\\\= p^{-20+18}q^{-16+12} \\\\= p^{-2}q^{-4} .\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}{p^2q^4} .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.