# How do you evaluate (\frac { x ^ { - 1/ 4} x ^ { 1/ 6} } { x ^ { 1/ 4} x ^ { - 1/ 2} } ) ^ { - 1/ 3}?

Sep 15, 2017

${x}^{- \frac{1}{18}} = \sqrt[18]{x}$

#### Explanation:

Remember that ${x}^{a} \cdot {x}^{b} = {x}^{a + b}$

...you can apply this to the numerator and denominator, giving:

${\left(\frac{{x}^{\frac{1}{6} - \frac{1}{4}}}{{x}^{\frac{1}{4} - \frac{1}{2}}}\right)}^{- \frac{1}{3}}$

$= {\left({x}^{- \frac{1}{12}} / \left({x}^{- \frac{1}{4}}\right)\right)}^{- \frac{1}{3}}$

Next, remember that $\frac{1}{x} ^ a = {x}^{- a}$

...and furthermore, $\frac{1}{x} ^ \left(- a\right) = \frac{1}{\frac{1}{x} ^ \left(a\right)} = {x}^{a}$

So with this in mind, you can flip your numerator/denominator upside down, and write:

$= {\left({x}^{\frac{1}{4}} / {x}^{\frac{1}{12}}\right)}^{- \frac{1}{3}}$

...and, since ${x}^{a} / {x}^{b} = {x}^{a - b}$, you can rewrite this as:

${\left({x}^{\frac{1}{4} - \frac{1}{12}}\right)}^{- \frac{1}{3}}$

$= {\left({x}^{\frac{1}{6}}\right)}^{- \frac{1}{3}}$

...and furthermore, remember that ${\left({x}^{a}\right)}^{b} = {x}^{a \cdot b}$

So you can rewrite it as:

$= {x}^{\frac{1}{6} \cdot - \frac{1}{3}} = {x}^{- \frac{1}{18}}$

...alternately, this could be written as $\frac{1}{\sqrt[18]{x}}$

...GOOD LUCK!