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

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

$\displaystyle \beta Px$ = $\displaystyle \alpha Py$

thanks

Printable View

- Nov 9th 2012, 11:26 AMentrepreneurforum.co.ukHelp simplifying this equation
$\displaystyle [( \beta / \alpha) (px / py)]^\beta = [(\alpha / \beta (py / px)]^\alpha$

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

$\displaystyle \beta Px$ = $\displaystyle \alpha Py$

thanks - Nov 9th 2012, 11:29 AMPlatoRe: Help simplifying this equation
- Nov 9th 2012, 12:07 PMskeeterRe: Help simplifying this equation
$\displaystyle \left(\frac{b}{a}\right)^b \left(\frac{px}{py}\right)^b = \left(\frac{a}{b}\right)^a \left(\frac{py}{px}\right)^a$

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

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

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

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

$\displaystyle bpx = apy$ - Nov 9th 2012, 01:00 PMentrepreneurforum.co.ukRe: Help simplifying this equation
Thanks, that was much clearer,

I have another question as well now :)

I have my Lagrangian functions

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

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

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

If they are, I've solved for Lambda

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

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

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