How do you simplify $\frac{3 x^{- 4} y^{5}}{{(2 x^{3} y^{- 7})}^{- 2}}$ $(3x^-4y^5)/(2x^3y^-7)^-2$ using only positive exponents?

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How do you simplify $\frac{3 x^{- 4} y^{5}}{{(2 x^{3} y^{- 7})}^{- 2}}$ $(3x^-4y^5)/(2x^3y^-7)^-2$ using only positive exponents?

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Answer 1 · 2016-03-22T18:01:36

order of operations requires that we deal with the exponent in the denominator first using the power to power rule.
this means that our expression now becomes
$\frac{3 x^{- 4} y^{5}}{2^{- 2} x^{- 6} y^{14}}$ $(3x^-4 y^5)/(2^-2x^-6y^14)$

Now we can transpose the factors with negative exponents to the opposite side of the fraction bar to get:
$\frac{3 (2^{2}) x^{6} y^{5}}{x^{4} y^{14}}$ $(3(2^2) x^6 y^5)/(x^4y^14)$

which now makes everything simple by using the subtraction rule for exponents when we divide with the same base.

$12 x^{2} y^{- 9}$ $12x^2y^-9$

which is finally simplified to

$\frac{12 x^{2}}{y^{9}}$ $(12 x^2)/(y^9)$

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How do you simplify $\frac{3 x^{- 4} y^{5}}{{(2 x^{3} y^{- 7})}^{- 2}}$ $(3x^-4y^5)/(2x^3y^-7)^-2$ using only positive exponents?

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How do you simplify $\frac{3 x^{- 4} y^{5}}{{(2 x^{3} y^{- 7})}^{- 2}}$ $(3x^-4y^5)/(2x^3y^-7)^-2$ using only positive exponents?