## Intermediate Algebra: Connecting Concepts through Application

$\dfrac{1029x^{8}z^{10}}{250y^{11}}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\left( \dfrac{2}{3}xy^5z^{-4} \right)^{-1}\left( \dfrac{5}{7}x^{-3}y^{2}z^{-2} \right)^{-3} ,$ use the laws of exponents. $\bf{\text{Solution Details:}}$ Using the Power of a Quotient Rule of the laws of exponents, which is given by $\left( \dfrac{x^m}{y^n} \right)^p=\dfrac{x^{mp}}{y^{np}},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \left( \dfrac{2}{3}xy^5z^{-4} \right)^{-1}\left( \dfrac{5}{7}x^{-3}y^{2}z^{-2} \right)^{-3} \\\\= \left( \dfrac{2xy^5z^{-4}}{3} \right)^{-1}\left( \dfrac{5x^{-3}y^{2}z^{-2}}{7} \right)^{-3} \\\\= \left( \dfrac{2^{-1}x^{-1}y^{5(-1)}z^{-4(-1)}}{3^{-1}} \right)\left( \dfrac{5^{-3}x^{-3(-3)}y^{2(-3)}z^{-2(-3)}}{7^{-3}} \right) \\\\= \left( \dfrac{2^{-1}x^{-1}y^{-5}z^{4}}{3^{-1}} \right)\left( \dfrac{5^{-3}x^{9}y^{-6}z^{6}}{7^{-3}} \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( \dfrac{2^{-1}x^{-1}y^{-5}z^{4}}{3^{-1}} \right)\left( \dfrac{5^{-3}x^{9}y^{-6}z^{6}}{7^{-3}} \right) \\\\= \dfrac{2^{-1}\left(5^{-3}\right)x^{-1+9}y^{-5+(-6)}z^{4+6}}{3^{-1}\left(7^{-3}\right)} \\\\= \dfrac{2^{-1}\left(5^{-3}\right)x^{8}y^{-11}z^{10}}{3^{-1}\left(7^{-3}\right)} .\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{2^{-1}\left(5^{-3}\right)x^{8}y^{-11}z^{10}}{3^{-1}\left(7^{-3}\right)} \\\\= \dfrac{3^{1}\left(7^{3}\right)x^{8}z^{10}}{2^{1}\left(5^{3}\right)y^{11}} \\\\= \dfrac{1029x^{8}z^{10}}{250y^{11}} .\end{array}