Question

Use the law of exponents to rewrite and simplify the expression.

$\frac{\sqrt{\text{a}\sqrt{\text{b}}}}{\sqrt[3]{\text{ab}}}$

Answer 1

To simplify the given expression:

Using the law of exponents to simplify the given expression in below:

$\begin{array}{rcl}\frac{\sqrt{\text{a}\sqrt{\text{b}}}}{\sqrt[3]{\text{ab}}}& =& \frac{{\left(\text{a}\sqrt{\text{b}}\right)}^{{\displaystyle \frac{1}{2}}}}{{\left(\text{ab}\right)}^{{\displaystyle \frac{1}{3}}}}\left[\text{Fraction exponent rule:}\left(\sqrt[\text{n}]{{\text{a}}^{\text{m}}}\right)={\text{a}}^{\frac{\text{m}}{\text{n}}}\right]\\ & \Rightarrow & \frac{\sqrt{\text{a}\sqrt{\text{b}}}}{\sqrt[3]{\text{ab}}}=\frac{{\text{a}}^{{\displaystyle \frac{1}{2}}}{\left(\sqrt{\text{b}}\right)}^{{\displaystyle \frac{1}{2}}}}{{\text{a}}^{{\displaystyle \frac{1}{3}}}{\text{b}}^{{\displaystyle \frac{1}{3}}}}\left[\text{Power of a product rule}:{\left(\text{ab}\right)}^{\text{m}}={\text{a}}^{\text{m}}{\text{×b}}^{\text{m}}\right]\\ & \Rightarrow & \frac{\sqrt{\text{a}\sqrt{\text{b}}}}{\sqrt[3]{\text{ab}}}=\frac{{\text{a}}^{{\displaystyle \frac{1}{2}}}{\left({\text{b}}^{{\displaystyle \frac{1}{2}}}\right)}^{{\displaystyle \frac{1}{2}}}}{{\text{a}}^{{\displaystyle \frac{1}{3}}}{\text{b}}^{{\displaystyle \frac{1}{3}}}}\left[\text{Simplifying}\right]\\ & \Rightarrow & \frac{\sqrt{\text{a}\sqrt{\text{b}}}}{\sqrt[3]{\text{ab}}}=\frac{{\text{a}}^{{\displaystyle \frac{1}{2}}}{\text{b}}^{{\displaystyle \frac{1}{4}}}}{{\text{a}}^{{\displaystyle \frac{1}{3}}}{\text{b}}^{{\displaystyle \frac{1}{3}}}}\left[\text{Power of a power rule}:{\left({\text{a}}^{\text{m}}\right)}^{\text{n}}={\text{a}}^{\text{m×n}}\right]\\ & \Rightarrow & \frac{\sqrt{\text{a}\sqrt{\text{b}}}}{\sqrt[3]{\text{ab}}}={\text{a}}^{\frac{1}{2}\text{-}\frac{1}{3}}{\text{b}}^{\frac{1}{4}\text{-}\frac{1}{3}}\left[\text{Quotient rule:}\frac{{\text{a}}^{\text{m}}}{{\text{a}}^{\text{n}}}{\text{=a}}^{\text{m-n}}\right]\\ \therefore \frac{\sqrt{\text{a}\sqrt{\text{b}}}}{\sqrt[3]{\text{ab}}}& =& {\text{a}}^{\frac{1}{6}}{\text{b}}^{\text{-}\frac{1}{12}}\left[\text{Simplifying}\right]\end{array}$

Hence, the answer is $\begin{array}{l}{\mathbf{\text{a}}}^{\frac{\mathbf{1}}{\mathbf{6}}}{\mathbf{\text{b}}}^{\mathbf{\text{-}}\frac{\mathbf{1}}{\mathbf{12}}}\end{array}$.