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Math Help - Solving an equation with logarithms in it.

  1. #1
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    Solving an equation with logarithms in it.

    how do I solve x in

    4+7 \ln e^{x^2} = 2 e^{4 \ln x}
    Last edited by mr fantastic; July 18th 2009 at 02:52 AM. Reason: Re-titled, added latex
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  2. #2
    MHF Contributor red_dog's Avatar
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    Use these identities: e^{\ln x}=\ln e^x=x

    Then, \ln e^{x^2}=x^2 and e^{4\ln x}=e^{\ln x^4}=x^4

    The equation becomes 4+7x^2=2x^4, \ x>0

    Let x^2=t, \ t>0\Rightarrow 2t^2-7t-4=0\Rightarrow t_1=4, \ t_2=-\frac{1}{2}

    Now replace t with 4: x^2=4\Rightarrow x=2
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