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.

2 Answers
Feb 20, 2018

#((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)#

Feb 20, 2018

# 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)#