## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 3 - Exponents, Polynomials and Functions - 3.1 Rules for Exponents - 3.1 Exercises - Page 234: 104

#### Answer

$3,125x^{10}y^{35}z^{5}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $(5x^2y^7z)^{\frac{3}{2}}(5x^2y^7z)^{\frac{7}{2}} ,$ use the laws of exponents. $\bf{\text{Solution Details:}}$ Using the Power of the Product 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} (5x^2y^7z)^{\frac{3}{2}}(5x^2y^7z)^{\frac{7}{2}} \\\\= \left( 5^{\frac{3}{2}}x^{2\cdot\frac{3}{2}}y^{7\cdot\frac{3}{2}}z^{\frac{3}{2}} \right) \left( 5^{\frac{7}{2}}x^{2\cdot\frac{7}{2}}y^{7\cdot\frac{7}{2}}z^{\frac{7}{2}} \right) \\\\= \left( 5^{\frac{3}{2}}x^{3}y^{\frac{21}{2}}z^{\frac{3}{2}} \right) \left( 5^{\frac{7}{2}}x^{7}y^{\frac{49}{2}}z^{\frac{7}{2}} \right) .\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} \left( 5^{\frac{3}{2}}x^{3}y^{\frac{21}{2}}z^{\frac{3}{2}} \right) \left( 5^{\frac{7}{2}}x^{7}y^{\frac{49}{2}}z^{\frac{7}{2}} \right) \\\\= 5^{\frac{3}{2}+\frac{7}{2}}x^{3+7}y^{\frac{21}{2}+\frac{49}{2}}z^{\frac{3}{2}+\frac{7}{2}} \\\\= 5^{\frac{10}{2}}x^{10}y^{\frac{70}{2}}z^{\frac{10}{2}} \\\\= 5^{5}x^{10}y^{35}z^{5} \\\\= 3,125x^{10}y^{35}z^{5} .\end{array}

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