College Algebra (11th Edition)

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

Chapter R - Review Exercises: 100

Answer

$\dfrac{n^{\frac{1}{2}}}{5m^{\frac{5}{2}}} $

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $ \left( \dfrac{25m^3n^5}{m^{-2}n^6} \right)^{-1/2} .$ $\bf{\text{Solution Details:}}$ 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} \left( 25m^{3-(-2)}n^{5-6} \right)^{-1/2} \\\\= \left( 25m^{3+2}n^{5-6} \right)^{-1/2} \\\\= \left( 25m^{5}n^{-1} \right)^{-1/2} .\end{array} 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} 25^{-\frac{1}{2}}m^{5\left(-\frac{1}{2} \right)}n^{-1\left(-\frac{1}{2} \right)} \\\\= 25^{-\frac{1}{2}}m^{-\frac{5}{2}}n^{\frac{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{n^{\frac{1}{2}}}{25^{\frac{1}{2}}m^{\frac{5}{2}}} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{n^{\frac{1}{2}}}{\sqrt{25^1}m^{\frac{5}{2}}} \\\\= \dfrac{n^{\frac{1}{2}}}{5m^{\frac{5}{2}}} .\end{array}
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