## Intermediate Algebra: Connecting Concepts through Application

$\dfrac{7x^{2}}{5y^{3}}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\left( \dfrac{49x^5y^3}{25xy^9} \right)^{\frac{1}{2}} ,$ 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{49x^5y^3}{25xy^9} \right)^{\frac{1}{2}} \\\\ \left( \dfrac{49x^{5-1}y^{3-9}}{25} \right)^{\frac{1}{2}} \\\\ \left( \dfrac{49x^{4}y^{-6}}{25} \right)^{\frac{1}{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{7^2x^{4}y^{-6}}{5^2} \right)^{\frac{1}{2}} \\\\= \dfrac{7^{2\cdot\frac{1}{2}}x^{4\cdot\frac{1}{2}}y^{-6\cdot\frac{1}{2}}}{5^{2\cdot\frac{1}{2}}} \\\\= \dfrac{7^{1}x^{2}y^{-3}}{5^{1}} \\\\= \dfrac{7x^{2}y^{-3}}{5} .\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{7x^{2}y^{-3}}{5} \\\\= \dfrac{7x^{2}}{5y^{3}} .\end{array}