How do you get rid of negative exponents in ${(\frac{x^{4} y^{- 2}}{x^{- 3} y^{5}})}^{- 1}$ $(\frac { x ^ { 4} y ^ { - 2} } { x ^ { - 3} y ^ { 5} } ) ^ { - 1}$ ?

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How do you get rid of negative exponents in ${(\frac{x^{4} y^{- 2}}{x^{- 3} y^{5}})}^{- 1}$ $(\frac { x ^ { 4} y ^ { - 2} } { x ^ { - 3} y ^ { 5} } ) ^ { - 1}$ ?

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Answer 1 · 2017-03-15T03:06:35

$\frac{y^{7}}{x^{7}}$ $y^7/x^7$

Explanation:

First, the -1 as the exponent of the whole fraction can simplify down to the reciprocal of the fraction itself.
$\frac{x^{- 3} y^{5}}{x^{4} y^{- 2}}$ $(x^-3y^5)/(x^4y^-2)$
then you can subtract the exponents in the denominator from the exponents in the numerator. This will give you
$x^{- 7} y^{7}$ $x^-7y^7$
$x^{- 7} = \frac{1}{x^{7}}$ $x^-7=1/x^7$ so
$(\frac{1}{x^{7}}) y^{7}$ $(1/x^7)y^7$
$\frac{y^{7}}{x^{7}}$ $y^7/x^7$

Answer 2 · 2017-03-15T04:02:17

$\frac{y^{7}}{x^{7}}$ $y^7/x^7$

Explanation:

Given: ${(\frac{x^{4} y^{- 2}}{x^{- 3} y^{5}})}^{- 1}$ $((x^4y^-2)/(x^-3y^5))^-1$

The key to this one is understanding that a negative exponent means division so the result will be a reciprocal. That means to flip the numerator 'top' and denominator 'bottom'.

Since the exponent outside the bracket is $- 1$ $-1$ we can clear it right away by flipping the whole equation to end up with a new one with exponent $1$ $1$ which means multiplied by $1$ $1$ :

${(\frac{x^{4} y^{- 2}}{x^{- 3} y^{5}})}^{- 1} = 1 \cdot \frac{x^{- 3} y^{5}}{x^{4} y^{- 2}}$ $((x^4y^-2)/(x^-3y^5))^-1 = 1* (x^-3y^5)/(x^4y^-2)$

But $x^{- 3} = \frac{1}{x^{3}}$ $x^-3 = 1/x^3$ ; and $y^{- 2} = \frac{1}{y^{2}}$ $y^-2 = 1/y^2$ ; so:

$\frac{x^{- 3} y^{5}}{x^{4} y^{- 2}} = \frac{\frac{y^{5}}{x^{3}}}{\frac{x^{4}}{y^{2}}}$ $(x^-3y^5)/(x^4y^-2) = (y^5/x^3)/(x^4/y^2)$ which we will need to invert and multiply:

$\frac{x^{- 3} y^{5}}{x^{4} y^{- 2}} = (\frac{y^{5}}{x^{3}}) \cdot (\frac{y^{2}}{x^{4}}) = \frac{y^{7}}{x^{7}}$ $(x^-3y^5)/(x^4y^-2) = (y^5/x^3)*(y^2/x^4) = y^7/x^7$

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How do you get rid of negative exponents in ${(\frac{x^{4} y^{- 2}}{x^{- 3} y^{5}})}^{- 1}$ $(\frac { x ^ { 4} y ^ { - 2} } { x ^ { - 3} y ^ { 5} } ) ^ { - 1}$ ?

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How do you get rid of negative exponents in (\frac { x ^ { 4} y ^ { - 2} } { x ^ { - 3} y ^ { 5} } ) ^ { - 1}(x4y−2x−3y5)−1(\frac { x ^ { 4} y ^ { - 2} } { x ^ { - 3} y ^ { 5} } ) ^ { - 1}?

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Explanation:

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How do you get rid of negative exponents in ${(\frac{x^{4} y^{- 2}}{x^{- 3} y^{5}})}^{- 1}$ $(\frac { x ^ { 4} y ^ { - 2} } { x ^ { - 3} y ^ { 5} } ) ^ { - 1}$ ?