1. ## exponent problems

Is it possible that $\displaystyle Y=A(tY)^{1-\alpha}K^{\alpha}$ is equivalent to $\displaystyle Y=A^\frac{1}{\alpha}t^\frac{1-\alpha}{\alpha}K$ ? Where the heck did the $\displaystyle Y$ that was on the right hand side go??

2. Originally Posted by garymarkhov
Is it possible that $\displaystyle Y=A(tY)^{1-\alpha}K^{\alpha}$ is equivalent to $\displaystyle Y=A^\frac{1}{\alpha}t^\frac{1-\alpha}{\alpha}K$ ? Where the heck did the $\displaystyle Y$ that was on the right hand side go??
$\displaystyle Y = At^{1-\alpha}Y^{1-\alpha}K^{\alpha}$

$\displaystyle \frac{Y}{Y^{1-\alpha}} = At^{1-\alpha}K^{\alpha}$

$\displaystyle Y^{\alpha} = At^{1-\alpha}K^{\alpha}$

$\displaystyle (Y^{\alpha})^{\frac{1}{\alpha}} = (At^{1-\alpha}K^{\alpha})^{\frac{1}{\alpha}}$

$\displaystyle Y=A^\frac{1}{\alpha}t^\frac{1-\alpha}{\alpha}K$

3. Originally Posted by skeeter
$\displaystyle Y = At^{1-\alpha}Y^{1-\alpha}K^{\alpha}$

$\displaystyle \frac{Y}{Y^{1-\alpha}} = At^{1-\alpha}K^{\alpha}$

$\displaystyle Y^{\alpha} = At^{1-\alpha}K^{\alpha}$

$\displaystyle (Y^{\alpha})^{\frac{1}{\alpha}} = (At^{1-\alpha}K^{\alpha})^{\frac{1}{\alpha}}$

$\displaystyle Y=A^\frac{1}{\alpha}t^\frac{1-\alpha}{\alpha}K$
That's fantastic.