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Math Help - more logs

  1. #1
    Newbie
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    Jan 2009
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    Smile more logs

    hiya
    well ive been given the formula (y/x)^a=be^-x cant make this end bit look right its to the power of -x
    and I have to make it a linear formula...was thinking I'm probably going to have to take logs but was wondering if I should take ln e so I can eliminate the e out

    OR does anyone know a better way to make this a linear formula

    much help appriciated
    Last edited by p3achy; January 7th 2009 at 12:05 PM. Reason: make formula look right
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  2. #2
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    Quote Originally Posted by p3achy View Post
    hiya
    well ive been given the formula (y/x)^a=be^-x cant make this end bit look right its to the power of -x
    and I have to make it a linear formula...was thinking I'm probably going to have to take logs but was wondering if I should take ln e so I can eliminate the e out

    OR does anyone know a better way to make this a linear formula

    much help appriciated
    Take logs!

     \ln{|\frac{y}{x}^a|} = \ln{|be^{-x}|}


     \ln{|\frac{y^a}{x^2}|} = \ln{|b|}+\ln{|e^{-x}|}

     \ln{|y^a|}-\ln{|x^2|} = \ln{|b|}+\ln{|e^{-x}|}

     \ln{|y^a|}-\ln{|x^2|} = \ln{|b|}-x

     a\ln{|y|}-ln(x^2) = \ln{|b|}-x

     a\ln{|y|} = \ln{|x^2|}  + \ln{|b|}-x

     a\ln{|y|} = \ln{|\frac{x^2}{b}|}-x

     \ln{|y|} = \frac{\ln{|\frac{x^2}{b}|}-x}{a}


     \ln{|y|} = \frac{1}{a}\times \ln{|\frac{x^2}{b}|}-\frac{x}{a}

    \displaystyle  \ln{|y|} =  \ln{|(\frac{x^2}{b})^{\frac{1}{a}}|}-\frac{x}{a}

     \displaystyle y =  e^{\ln{|(\frac{x^2}{b})^{\frac{1}{a}}|}-\frac{x}{a}}

    \displaystyle y =  (\frac{x^2}{b})^{\frac{1}{a}}.\frac{1}{e^{\frac{x}  {a}}}
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  3. #3
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    thx looks like i was mostly there just messed me logs up at the end
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