Help simplifying this equation

**entrepreneurforum.co.uk** · November 9th 2012, 11:26 AM

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

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

$\beta Px$ = $\alpha Py$

thanks

**Plato** · November 9th 2012, 11:29 AM

Originally Posted by entrepreneurforum.co.uk

((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.

**skeeter** · November 9th 2012, 12:07 PM

Originally Posted by entrepreneurforum.co.uk

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

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

$\beta Px$ = $\alpha Py$

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

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

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

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

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

$bpx = apy$

**entrepreneurforum.co.uk** · November 9th 2012, 01:00 PM

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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