Intermediate Algebra (12th Edition)

$\dfrac{1}{m^{1/4}n^{3/4}}$
$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $\dfrac{m^{3/4}n^{-1/4}}{\left( m^2n \right)^{1/2}} .$ $\bf{\text{Solution Details:}}$ 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} \dfrac{m^{3/4}n^{-1/4}}{m^{2\cdot\frac{1}{2}}n^{\frac{1}{2}}} \\\\= \dfrac{m^{3/4}n^{-1/4}}{m^{1}n^{\frac{1}{2}}} \\\\= \dfrac{m^{3/4}n^{-1/4}}{mn^{\frac{1}{2}}} .\end{array} 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} m^{\frac{3}{4}-1}n^{-\frac{1}{4}-\frac{1}{2}} \\\\= m^{\frac{3}{4}-\frac{4}{4}}n^{-\frac{1}{4}-\frac{2}{4}} \\\\= m^{-\frac{1}{4}}n^{-\frac{3}{4}} .\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{1}{m^{\frac{1}{4}}n^{\frac{3}{4}}} \\\\= \dfrac{1}{m^{1/4}n^{3/4}} .\end{array}