How do you simplify $\frac{a^{- 2} b^{- 3}}{c^{- 1}}$ $\frac{a^{-2}b^{-3}}{c^{-1}}$ without any negative or fractional exponents in the answer?

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How do you simplify $\frac{a^{- 2} b^{- 3}}{c^{- 1}}$ $\frac{a^{-2}b^{-3}}{c^{-1}}$ without any negative or fractional exponents in the answer?

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Answer 1 · 2014-12-19T16:44:12

Well, I am not completely sure it is what you want but here it is: $\frac{c}{a^{2} b^{3}}$ $c/(a^2b^3)$

what happens is that a negative exponent allows you to "change floor" to your factor in a fraction.
So if you have:
$a^{- 2}$ $a^(-2)$ it can go "downstairs" to the denominator loosing the minus sign: $\frac{1}{a^{2}}$ $1/(a^2)$ ;
Same thing with $b^{- 3}$ $b^(-3)$ that becomes: $\frac{1}{b^{3}}$ $1/(b^3)$ ;
If the factor is already "downstairs", in the denominator, you can take it "upstairs" to the numerator, loosing again the minus sign.
$\frac{1}{c^{- 1}}$ $1/(c^(-1)$ becomes: $c^{1} = c$ $c^1=c$ .

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How do you simplify $\frac{a^{- 2} b^{- 3}}{c^{- 1}}$ $\frac{a^{-2}b^{-3}}{c^{-1}}$ without any negative or fractional exponents in the answer?

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