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Math Help - negative exponents

  1. #1
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    Question negative exponents

    I am having a problem with manipulating negaticve exponents

    the problem is 1/2x^-2y^-3


    I come up with 2x^2y^3 but that does not seem right.

    Any thoughts


    Thanks

    JS
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jsto View Post
    I am having a problem with manipulating negaticve exponents

    the problem is 1/2x^-2y^-3


    I come up with 2x^2y^3 but that does not seem right.

    Any thoughts


    Thanks

    JS
    Just making sure...is it \frac{1}{2x^{-2}y^{-3}}??
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Just making sure...is it \frac{1}{2x^{-2}y^{-3}}??
    Yes that's right
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  4. #4
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    EDIT: Cocked it up, see correct answer below
    Last edited by rowe; October 12th 2009 at 11:07 AM.
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  5. #5
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    Quote Originally Posted by rowe View Post
    \frac{1}{2x^{-2}y^{-3}} = \frac{1}{\frac{1}{2x^2y^3}}=1 \cdot \frac{2x^2y^3}{1} = 2x^2y^3
    I don't want to pick at you but the 2 in the denominator has NOT a negative exponent

    \frac{1}{2x^{-2}y^{-3}} = \frac{1}{2 \cdot \frac{1}{x^2y^3}}= \dfrac{x^2y^3}2
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  6. #6
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by rowe View Post
    \frac{1}{2x^{-2}y^{-3}} = \frac{1}{\frac{1}{2x^2y^3}}=1 \cdot \frac{2x^2y^3}{1} = 2x^2y^3
    The 2 is not exponentiated so you're a factor of 4 too high.

    \frac{1}{2}x^2y^3
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