## Intermediate Algebra (12th Edition)

$\dfrac{1}{b^{19/12}}$
$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $\dfrac{ \left( a^{2}b^{5} \right)^{-1/4}}{\left( a^{-3}b^2 \right)^{1/6}} .$ $\bf{\text{Solution Details:}}$ 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{ a^{2\cdot\left(-\frac{1}{4} \right)}b^{5\cdot\left(-\frac{1}{4} \right)}}{a^{-3\cdot\frac{1}{6}}b^{2\cdot\frac{1}{6}}} \\\\= \dfrac{ a^{-\frac{2}{4}}b^{-\frac{5}{4}}}{a^{-\frac{3}{6}}b^{\frac{2}{6}}} \\\\= \dfrac{ a^{-\frac{1}{2}}b^{-\frac{5}{4}}}{a^{-\frac{1}{2}}b^{\frac{1}{3}}} \\\\= \dfrac{ \cancel{a^{-\frac{1}{2}}}b^{-\frac{5}{4}}}{\cancel{a^{-\frac{1}{2}}}b^{\frac{1}{3}}} \\\\= \dfrac{ b^{-\frac{5}{4}}}{b^{\frac{1}{3}}} .\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} b^{-\frac{5}{4}-\frac{1}{3}} .\end{array} To simplify the expression $-\dfrac{5}{4}-\dfrac{1}{3} ,$ change the expressions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $4$ and $3$ is $12$ since it is the lowest number that can be exactly divided by the denominators. Multiplying the terms by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to \begin{array}{l}\require{cancel} b^{-\frac{5}{4}\cdot\frac{3}{3}-\frac{1}{3}\cdot\frac{4}{4}} \\\\= b^{-\frac{15}{12}-\frac{4}{12}} \\\\= b^{-\frac{19}{12}} .\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}{b^{\frac{19}{12}}} \\\\= \dfrac{1}{b^{19/12}} .\end{array}