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

Question

1246 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-15T13:58:48

$x^(-1/18) = root(18)x$

Remember that $x^a * x^b = x^(a + b)$

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

$((x^(1/6-1/4))/(x^(1/4 - 1/2)))^(-1/3)$

$=(x^(-1/12)/(x^(-1/4)))^(-1/3)$

Next, remember that $1/x^a = x^(-a)$

...and furthermore, $1/x^(-a) = 1/(1/x^(a)) = x^a$

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

$= (x^(1/4)/x^(1/12))^(-1/3)$

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

$(x^(1/4 - 1/12))^(-1/3)$

$= (x^(1/6))^(-1/3)$

...and furthermore, remember that $(x^a)^b = x^(a * b)$

So you can rewrite it as:

$= x^(1/6 * -1/3) = x^(-1/18)$

...alternately, this could be written as $1/root(18)x$

...GOOD LUCK!

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