# How do you simplify (sqrt(x) / (y^-2)) ^4 ?

Apr 4, 2017

${\left(\frac{\sqrt{x}}{y} ^ - 2\right)}^{4} = {x}^{2} {y}^{8}$

#### Explanation:

To start we can deal with the negative exponent using the rule ${a}^{-} b = \frac{1}{a} ^ b$. Then:

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

We should rewrite $\sqrt{x}$ with a fractional exponent:

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

Split up the exponent:

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

Use the rule ${\left({a}^{b}\right)}^{c} = {a}^{b c}$:

${\left({y}^{2}\right)}^{4} {\left({x}^{\frac{1}{2}}\right)}^{4} = {y}^{2 \left(4\right)} {x}^{\frac{1}{2} \left(4\right)} = {x}^{2} {y}^{8}$