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.
I have a test on this tomorrow, so I would appreciate a thurough explanation.
2 Answers
Explanation:
Note that if
#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
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)#
Explanation:
There are several laws of indices, but none is more important than another, so you apply them in any order.
A useful law is:
Notice that in the fraction we are given, the index is negative.
Let's get rid of the negative.
Recall the law
Let's get rid of all the negative indices with this law.
Recall:
Recall:
Recall: