Question

Solve for the exponent of x?

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

I have a test on this tomorrow, so I would appreciate a thurough explanation.

Answer 1

`((x^(-1/3) x^(1/6))/(x^(1/4) x^(-1/2)))^(-1/3) = x^(-1/36)`

Explanation:

Note that if `x > 0` then:

`x^a x^b = x^(a+b)`

Also:

`x^(-a) = 1/x^a`

Also:

`(x^a)^b = x^(ab)`

In the given example, we might as well assume `x > 0` since otherwise we are faced with non-real values for `x < 0` and undefined value for `x = 0`.

So we find:

`((x^(-1/3) x^(1/6))/(x^(1/4) x^(-1/2)))^(-1/3) = ((x^(-1/3 +1/6))/(x^(1/4 - 1/2)))^(-1/3)`

`color(white)(((x^(-1/3) x^(1/6))/(x^(1/4) x^(-1/2)))^(-1/3)) = ((x^(-1/6))/(x^(-1/4)))^(-1/3)`

`color(white)(((x^(-1/3) x^(1/6))/(x^(1/4) x^(-1/2)))^(-1/3)) = (x^(1/4) x^(-1/6))^(-1/3)`

`color(white)(((x^(-1/3) x^(1/6))/(x^(1/4) x^(-1/2)))^(-1/3)) = (x^(1/4-1/6))^(-1/3)`

`color(white)(((x^(-1/3) x^(1/6))/(x^(1/4) x^(-1/2)))^(-1/3)) = (x^(1/12))^(-1/3)`

`color(white)(((x^(-1/3) x^(1/6))/(x^(1/4) x^(-1/2)))^(-1/3)) = x^(1/12*(-1/3))`

`color(white)(((x^(-1/3) x^(1/6))/(x^(1/4) x^(-1/2)))^(-1/3)) = x^(-1/36)`

Answer 2

`x^(-1/36)`

Explanation:

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

There are several laws of indices, but none is more important than another, so you apply them in any order.

A useful law is: `" " (a/b)^-m = (b/a)^m`

Notice that in the fraction we are given, the index is negative.
Let's get rid of the negative.

`(color(blue)(x^(-1/3)x^(1/6))/(x^(1/4)x^(-1/2)))^color(red)(-1/3)= ((x^(1/4)x^(-1/2))/(color(blue)(x^(-1/3)x^(1/6))))^color(red)(1/3)`

Recall the law `" "x^-m = 1/x^m" and " 1/x^-n = x^n`

Let's get rid of all the negative indices with this law.

`((x^(1/4)x^(1/3))/(x^(1/6)x^(1/2)))^(1/3)`

Recall: `" "x^m x^n = x^(m+n)" "larr` add the indices

`((x^(1/4)x^(1/3))/(x^(1/6)x^(1/2)))^(1/3)= (x^(7/12)/x^(4/6))^(1/3)`

Recall: `" "x^m/x^n= x^(m-n)" "larr` subtract the indices

`(x^(7/12)/x^(4/6))^(1/3) = (x^(7/12-8/12))^(1/3) = (x^(-1/12))^(1/3)`

Recall:`" "(x^m)^n = x^(mn)" "larr` multiply the indices

`(x^(-1/12))^(1/3) = x^(-1/36)`