## Intermediate Algebra (12th Edition)

$\dfrac{y}{18x}$
$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $(3x^2y^{-1})^{-2}(2x^{-3}y)^{-1} .$ $\bf{\text{Solution Details:}}$ 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} (3^{-2}x^{2(-2)}y^{-1(-2)})(2^{-1}x^{-3(-1)}y^{-1}) \\\\= (3^{-2}x^{-4}y^{2})(2^{-1}x^{3}y^{-1}) .\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} (3^{-2})(2^{-1})x^{-4+3}y^{2+(-1)} \\\\= (3^{-2})(2^{-1})x^{-4+3}y^{2-1} \\\\= (3^{-2})(2^{-1})x^{-1}y^{1} \\\\= (3^{-2})(2^{-1})x^{-1}y .\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{y}{(3^{2})(2^{1})x^{1}} \\\\= \dfrac{y}{(9)(2)x} \\\\= \dfrac{y}{18x} .\end{array}