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

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

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