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Math Help - simplification of exponential expression

  1. #1
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    simplification of exponential expression

    [(x^2 y^-2)^-1]^-1=
    (1/x^2 y^-2)^-1=
    (y^2/x^2)-1=
    x^2/y^2
    this solution is right according to the back of the book, are the steps also correct?
    Last edited by dymand68; April 1st 2010 at 06:05 PM.
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  2. #2
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    Quote Originally Posted by dymand68 View Post
    [(x^2y^-2)^-1]^-1=
    (1/x^2y^-2)^-1=
    (y^2/x^2)-1=
    x^2/y^2
    this solution is right according to the back of the book, are the steps also correct?
    [(x^2y^{-2})^{-1}]^{-1}

     = (x^2y^{-2})^1

     = x^2y^{-2}

     = \frac{x^2}{y^2}

     = \left(\frac{x}{y}\right)^2.
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  3. #3
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    ty.........so the correct order of operations starts with the negative exponent OUTSIDE the brackets?
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