What is #((x^(1/2)y^(-7/4))/(y^(-5/4)))^-4#?

1 Answer
Sep 4, 2017

#y^2/x^2#

Explanation:

Let's get rid of all the negative indices first:

Recall: #(a/b)^-m = (b/a)^m#

#((x^(1/2)y^(-7/4))/(y^(-5/4)))^color(blue)(-4)#

#= ((y^(-5/4))/(x^(1/2)y^(-7/4)))^color(blue)(4)" "larr#invert the fraction

Recall: #x^-m = 1/x^m and 1/x^-n = x^n#

#= ((y^(7/4))/(x^(1/2)y^(5/4)))^4#

Recall: # x^m/x^n = x^(m-n)#

#= ((y^(7/4 -5/4))/(x^(1/2)))^4#

#=(y^(1/2)/x^(1/2))^4color(white)(xxxxxxxxxxx)(y^(2/4) = y^(1/2))#

Multiply the indices

#=y^2/x^2#