How do you simplify $\frac{x^{- 8} {(x^{- 2} y^{2})}^{2}}{{(4 x^{- 4})}^{2} y^{2}}$ $\frac { x ^ { - 8} ( x ^ { - 2} y ^ { 2} ) ^ { 2} } { ( 4x ^ { - 4} ) ^ { 2} y ^ { 2} }$ ?

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How do you simplify $\frac{x^{- 8} {(x^{- 2} y^{2})}^{2}}{{(4 x^{- 4})}^{2} y^{2}}$ $\frac { x ^ { - 8} ( x ^ { - 2} y ^ { 2} ) ^ { 2} } { ( 4x ^ { - 4} ) ^ { 2} y ^ { 2} }$ ?

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Answer 1 · 2018-04-23T00:47:17

$\frac{x^{- 8} {(x^{- 2} y^{2})}^{2}}{{(4 x^{- 4})}^{2} y^{2}} = \frac{y^{2}}{16 x^{4}}$ $(x^-8(x^-2y^2)^2)/((4x^-4)^2y^2)=y^2/(16x^4)$

Explanation:

$\frac{x^{- 8} {(x^{- 2} y^{2})}^{2}}{{(4 x^{- 4})}^{2} y^{2}}$ $(x^-8(x^-2y^2)^2)/((4x^-4)^2y^2)$

Let's loose the parenthesis first. Remember that

${(x^{m})}^{n} = x^{m n}$ $(x^m)^n=x^(mn)$ .

$\frac{x^{- 8} {(x^{- 2} y^{2})}^{2}}{{(4 x^{- 4})}^{2} y^{2}} = \frac{x^{- 8} x^{- 4} y^{4}}{16 x^{- 8} y^{2}}$ $(x^-8(x^-2y^2)^2)/((4x^-4)^2y^2)= (x^-8x^-4y^4)/(16x^-8y^2)$

Now cancel. Remember that

$\frac{x^{m}}{x^{n}} = x^{m - n}$ $(x^m)/(x^n)=x^(m-n)$ .

$\frac{x^{- 8} x^{- 4} y^{4}}{16 x^{- 8} y^{2}} = \frac{x^{- 4} y^{2}}{16}$ $(x^-8x^-4y^4)/(16x^-8y^2)=(x^-4y^2)/16$

Now express everything with positive exponents. Remember that

$x^{- m} = \frac{1}{x^{m}}$ $x^-m=1/x^m$

$\frac{x^{- 4} y^{2}}{16} = \frac{y^{2}}{16 x^{4}}$ $(x^-4y^2)/16=y^2/(16x^4)$

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How do you simplify $\frac{x^{- 8} {(x^{- 2} y^{2})}^{2}}{{(4 x^{- 4})}^{2} y^{2}}$ $\frac { x ^ { - 8} ( x ^ { - 2} y ^ { 2} ) ^ { 2} } { ( 4x ^ { - 4} ) ^ { 2} y ^ { 2} }$ ?

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Explanation:

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How do you simplify x−8(x−2y2)2(4x−4)2y2\frac { x ^ { - 8} ( x ^ { - 2} y ^ { 2} ) ^ { 2} } { ( 4x ^ { - 4} ) ^ { 2} y ^ { 2} }?

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Explanation:

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How do you simplify $\frac{x^{- 8} {(x^{- 2} y^{2})}^{2}}{{(4 x^{- 4})}^{2} y^{2}}$ $\frac { x ^ { - 8} ( x ^ { - 2} y ^ { 2} ) ^ { 2} } { ( 4x ^ { - 4} ) ^ { 2} y ^ { 2} }$ ?