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

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Answer 1 · 2017-09-15T13:58:48

$x^{- \frac{1}{18}} = \sqrt[18]{x}$ $x^(-1/18) = root(18)x$

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

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

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

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

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

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

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

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

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

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

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

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

So you can rewrite it as:

$= x^{\frac{1}{6} \cdot - \frac{1}{3}} = x^{- \frac{1}{18}}$ $= x^(1/6 * -1/3) = x^(-1/18)$

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

...GOOD LUCK!

How do you evaluate (x−14x16x14x−12)−13(\frac { x ^ { - 1/ 4} x ^ { 1/ 6} } { x ^ { 1/ 4} x ^ { - 1/ 2} } ) ^ { - 1/ 3}?