# How do you simplify (x^-4y^3)^(1/8)/(x^2y^5)^(-1/4)?

Sep 8, 2015

${y}^{- \frac{7}{8}}$

#### Explanation:

Start by looking at the numerator

${\left({x}^{- 4} {y}^{3}\right)}^{\frac{1}{8}}$

You can use the power of a product and power of a power properties of exponents to write

${\left({x}^{- 4} {y}^{3}\right)}^{\frac{1}{8}} = {\left({x}^{- 4}\right)}^{\frac{1}{8}} \cdot {\left({y}^{3}\right)}^{\frac{1}{8}}$

$= {x}^{- 4 \cdot \frac{1}{8}} \cdot {y}^{3 \cdot \frac{1}{8}} = {x}^{- \frac{1}{2}} \cdot {y}^{\frac{3}{8}}$

Do the same for the denominator

${\left({x}^{2} {y}^{5}\right)}^{- \frac{1}{4}} = {\left({x}^{2}\right)}^{- \frac{1}{4}} \cdot {\left({y}^{5}\right)}^{- \frac{1}{4}}$

$= {x}^{2 \cdot \left(- \frac{1}{4}\right)} \cdot {y}^{5 \cdot \left(- \frac{1}{4}\right)} = {x}^{- \frac{1}{2}} \cdot {y}^{- \frac{5}{4}}$

The original expression can now be simplified to

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{- \frac{1}{2}}}}} \cdot {y}^{\frac{3}{8}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{- \frac{1}{2}}}}} \cdot {y}^{- \frac{5}{4}}}$

You know that

$\textcolor{b l u e}{{x}^{- n} = \frac{1}{x} ^ n} \text{ }$, provided that $\textcolor{b l u e}{x \ne 0}$.

This means that you can write

${y}^{- \frac{5}{4}} = \frac{1}{y} ^ \left(\frac{5}{4}\right) \text{ }$, with $y \ne 0$

The expression becomes

${y}^{\frac{3}{8}} \cdot {y}^{- \frac{5}{4}} = {y}^{\frac{3}{8} - \frac{5}{4}} = \textcolor{g r e e n}{{y}^{- \frac{7}{8}}}$