Thread: exponent problems

1. exponent problems

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

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

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

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

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

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

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

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

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

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

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