Intermediate Algebra (12th Edition)

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

Chapter 7 - Section 7.2 - Rational Exponents - 7.2 Exercises: 77

Answer

$p^{2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $ \dfrac{ p^{1/5}p^{7/10}p^{1/2}}{\left( p^{3} \right)^{-1/5}} .$ $\bf{\text{Solution Details:}}$ 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{ p^{1/5}p^{7/10}p^{1/2}}{p^{3\cdot\left(-\frac{1}{5} \right)}} \\\\= \dfrac{ p^{1/5}p^{7/10}p^{1/2}}{p^{-\frac{3}{5}}} .\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{ p^{\frac{1}{5}+\frac{7}{10}+\frac{1}{2}}}{p^{-\frac{3}{5}}} .\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^{\frac{1}{5}+\frac{7}{10}+\frac{1}{2}-\left(-\frac{3}{5}\right)} \\\\ p^{\frac{1}{5}+\frac{7}{10}+\frac{1}{2}+\frac{3}{5}} \\\\ p^{\frac{2}{10}+\frac{7}{10}+\frac{5}{10}+\frac{6}{10}} \\\\ p^{\frac{20}{10}} \\\\ p^{2} .\end{array}
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