## Intermediate Algebra: Connecting Concepts through Application

$\dfrac{25}{9c^{4}}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\left( \dfrac{3c^5d^2}{5c^3d^2} \right)^{-2} ,$ use the laws of exponents. $\bf{\text{Solution Details:}}$ 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( \dfrac{3c^5d^2}{5c^3d^2} \right)^{-2} \\\\ \left( \dfrac{3c^{5-3}d^{2-2}}{5} \right)^{-2} \\\\ \left( \dfrac{3c^{2}d^{0}}{5} \right)^{-2} \\\\ \left( \dfrac{3c^{2}(1)}{5} \right)^{-2} \\\\ \left( \dfrac{3c^{2}}{5} \right)^{-2} .\end{array} 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{3c^{2}}{5} \right)^{-2} \\\\= \dfrac{3^{-2}c^{2(-2)}}{5^{-2}} \\\\= \dfrac{3^{-2}c^{-4}}{5^{-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{3^{-2}c^{-4}}{5^{-2}} \\\\= \dfrac{5^{2}}{3^{2}c^{4}} \\\\= \dfrac{25}{9c^{4}} .\end{array}