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

Apr 11, 2018

$\frac{{x}^{\frac{17}{24}}}{y}$

#### Explanation:

Simplify: ${\left({x}^{- \frac{1}{2}} {y}^{4}\right)}^{\frac{1}{4}} / \left({x}^{\frac{2}{3}} {y}^{\frac{3}{2}} {x}^{- \frac{3}{2}} {y}^{\frac{1}{2}}\right)$

First, let's distribute the $\frac{1}{4}$ exponent:

$\frac{{x}^{{\left(- \frac{1}{2}\right)}^{\frac{1}{4}}} {y}^{{\left(4\right)}^{\frac{1}{4}}}}{{x}^{\frac{2}{3}} {y}^{\frac{3}{2}} {x}^{- \frac{3}{2}} {y}^{\frac{1}{2}}}$

When you raise a power to a power you multiply them to find the exponent. Let's do that

$\frac{{x}^{\left(- \frac{1}{2}\right) \left(\frac{1}{4}\right)} {y}^{\left(4\right) \left(\frac{1}{4}\right)}}{{x}^{\frac{2}{3}} {y}^{\frac{3}{2}} {x}^{- \frac{3}{2}} {y}^{\frac{1}{2}}}$

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

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

Now let's combine the exponents with like bases in the denominator. When you multiply exponents with like bases you add the exponents together

$\frac{{x}^{- \frac{1}{8}} y}{{x}^{\left(\frac{2}{3}\right) + \left(- \frac{3}{2}\right)} {y}^{\left(\frac{3}{2}\right) + \left(\frac{1}{2}\right)}}$

$\frac{{x}^{- \frac{1}{8}} y}{{x}^{\left(\frac{4}{6}\right) + \left(- \frac{9}{6}\right)} {y}^{\frac{4}{2}}}$

$\frac{{x}^{- \frac{1}{8}} y}{{x}^{- \frac{5}{6}} {y}^{2}}$

When there is a negative exponent in the numerator we can bring it to the denominator as a positive exponent and vice versa

Let's bring ${x}^{- \frac{5}{6}}$ to the numerator since it is bigger than ${x}^{- \frac{1}{8}}$ so when we multiply them and add the exponents it will be positive.

$\frac{{x}^{- \frac{1}{8}} {x}^{\frac{5}{6}} y}{{y}^{2}}$

$\frac{{x}^{\left(- \frac{1}{8}\right) + \left(\frac{5}{6}\right)} y}{{y}^{2}}$

$\frac{{x}^{\left(- \frac{3}{24}\right) + \left(\frac{20}{24}\right)} y}{{y}^{2}}$

$\frac{{x}^{\frac{17}{24}} y}{{y}^{2}}$

Now we can cancel out one of the $y$'s on top and bottom

$\frac{{x}^{\frac{17}{24}} \cancel{y}}{{y}^{\cancel{2} \textcolor{red}{1}}}$

$\frac{{x}^{\frac{17}{24}}}{y}$

$\frac{{x}^{\frac{17}{24}}}{y}$