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Math Help - solving for x (negative exponent)

  1. #1
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    solving for x (negative exponent)

    Hey!

    I need some help with solving this equation for x...

    (1+(6^2*x^-2))^-1*(-6x^-2)-(1+(4^2*x^-2))^-1*(-4x^-2) = 0

    I have problem with (1+(6^2*x^-2))^-1 and (1+(4^2*x^-2))^-1

    thanks
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  2. #2
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    (1+(6^2x^{-2}))^{-1}(-6x^{-2})-(1+(4^2x^{-2}))^{-1}(-4x^{-2}) = 0


    x^{-a} = \frac 1{x^a}


    \frac 1{(1+(6^2x^{-2}))} * \frac{-6}{x^{2}}-\frac 1{(1+(4^2x^{-2}))} * \frac{-4}{x^{2}} = 0

    \frac {-6}{x^{-2}(1+(6^2x^{-2}))} - \frac{-4}{x^{-2}(1 + (4^2x^{-2}))} = 0
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  3. #3
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    Quote Originally Posted by derfleurer View Post
    (1+(6^2x^{-2}))^{-1}(-6x^{-2})-(1+(4^2x^{-2}))^{-1}(-4x^{-2}) = 0


    x^{-a} = \frac 1{x^a}


    \frac 1{(1+(6^2x^{-2}))} * \frac{-6}{x^{2}}-\frac 1{(1+(4^2x^{-2}))} * \frac{-4}{x^{2}} = 0
    The two x^2 should be x^{-2} shouldn't they?

    \frac {-6}{x^{-2}(1+(6^2x^{-2}))} - \frac{-4}{x^{-2}(1 + (4^2x^{-2}))} = 0
    Okay now the they have become x^{-2}.
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  4. #4
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    Haha, ok, so I gutted that last line. =p

    \frac {-6}{x^2(1+(6^2x^{-2}))} - \frac{-4}{x^2(1 + (4^2x^{-2}))} = 0
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  5. #5
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    wow, quick responses

    Thank you so much guys. your answers helped me a lot.
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