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  1. #1
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    Help simplifying this equation

    [( \beta / \alpha) (px / py)]^\beta = [(\alpha / \beta (py / px)]^\alpha

    I have the solution but my simplifying isn't the best;

    \beta Px = \alpha Py

    thanks
    Last edited by entrepreneurforum.co.uk; November 9th 2012 at 11:55 AM.
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  2. #2
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    Re: Help simplifying this equation

    Quote Originally Posted by entrepreneurforum.co.uk View Post
    ((Beta / Alpha) (px / py))^beta = ((alpha / beta) (py / px))^alpha

    I have the solution but my simplifying isn't the best;

    BetaPx = AlphaPy
    That notation makes it impossible to read your post.
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  3. #3
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    Re: Help simplifying this equation

    Quote Originally Posted by entrepreneurforum.co.uk View Post
    [( \beta / \alpha) (px / py)]^\beta = [(\alpha / \beta (py / px)]^\alpha

    I have the solution but my simplifying isn't the best;

    \beta Px = \alpha Py
    \left(\frac{b}{a}\right)^b \left(\frac{px}{py}\right)^b = \left(\frac{a}{b}\right)^a \left(\frac{py}{px}\right)^a

    \left(\frac{b}{a}\right)^b \left(\frac{a}{b}\right)^{-a} = \left(\frac{px}{py}\right)^{-b} \left(\frac{py}{px}\right)^a

    \left(\frac{b}{a}\right)^b \left(\frac{b}{a}\right)^{a} = \left(\frac{py}{px}\right)^{b} \left( \frac{py}{px} \right)^a

    \left(\frac{b}{a}\right)^{a+b} = \left(\frac{py}{px}\right)^{a+b}

    \frac{b}{a} = \frac{py}{px}

    bpx = apy
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  4. #4
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    Re: Help simplifying this equation

    Thanks, that was much clearer,

    I have another question as well now

    I have my Lagrangian functions

    1) y^{1/2} + \lambda px

    2) (1/2) x^{-1/2} + \lambda px

    Are they correct? from the Cobb Douglas function xy^{1/2}

    If they are, I've solved for Lambda

    1)  \lambda = y^{1/2} / px

    2)  \lambda = (1/2) x^{-1/2} / py

    Are these correct? as I now have to put them equal to each other and solve for x and y but my answers look a little confusing

    thanks
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