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how do you solve this (3x1/2x8) to the power of -2
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|dw:1364840332590:dw|
\[(\frac{3x}{2x^8})^{-2}=(\frac{2x^8}{3x})^2=(\frac{2x^7}{3})^2=\frac{(2x^7)^2}{(3)^2}\]
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