How do you simplify #(x^-4y^3)^(1/8)/(x^2y^5)^(-1/4)#?

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Answer 1 · 2015-09-08T22:03:50

#y^(-7/8)#

Start by looking at the numerator

#(x^(-4)y^3)^(1/8)#

You can use the power of a product and power of a power properties of exponents to write

#(x^(-4)y^3)^(1/8) = (x^(-4))^(1/8) * (y^3)^(1/8)#

#=x^(-4 * 1/8) * y^(3 * 1/8) = x^(-1/2) * y^(3/8)#

Do the same for the denominator

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

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

The original expression can now be simplified to

#(color(red)(cancel(color(black)(x^(-1/2)))) * y^(3/8))/(color(red)(cancel(color(black)(x^(-1/2)))) * y^(-5/4))#

You know that

#color(blue)(x^(-n) = 1/x^n)" "#, provided that #color(blue)(x!=0)#.

This means that you can write

#y^(-5/4) = 1/y^(5/4)" "#, with #y!=0#

The expression becomes

#y^(3/8) * y^(-5/4) = y^(3/8 - 5/4) = color(green)(y^(-7/8))#