Reducing a lot exponentation So I need to show that the following equation is true:
$(\frac{α}{(1-α )}·\frac{p_2}{p_1})^{1-α}·u=(\frac{α}{p_1})·(p_1 (\frac{α}{(1-α )}·\frac{p_2}{p_1})^{1-α}+p_2 (\frac{1-α}{α}·\frac{p_1}{p_2} )^α )·u$
I have been trying a lot of stuff (mainly working/reducing/trying different things on the right hand site. But I just cannot seem to get it to match.
 A: Notice that $(\frac{α}{(1-α )}·\frac{p_2}{p_1})^{1-α} = (\frac{(1-α)}{α}·\frac{p_1}{p_2})^{\alpha - 1}$, so the RHS looks like $(\frac{α}{p_1})·(p_1 (\frac{(1-α)}{α}·\frac{p_1}{p_2})^{\alpha - 1}+p_2 (\frac{1-α}{α}·\frac{p_1}{p_2} )^α )·u$.
Now we can factor the RHS to be $(\frac{α}{p_1})·(\frac{(1-α)}{α}·\frac{p_1}{p_2})^{\alpha - 1}·(p_1 +p_2 (\frac{1-α}{α}·\frac{p_1}{p_2} )^1 )·u$
$ = (\frac{α}{p_1})·(\frac{(1-α)^{\alpha - 1}}{α^{\alpha - 1}}·\frac{(p_1)^{\alpha - 1}}{(p_2)^{\alpha - 1}})·(p_1 + (\frac{1-α}{α}·\frac{p_1}{1} ) )·u$
$= (\frac{(1-α)^{\alpha - 1}}{α^{\alpha - 2}}·\frac{(p_1)^{\alpha - 2}}{(p_2)^{\alpha - 1}})·(p_1 + (\frac{1-α}{α}·\frac{p_1}{1} ) )·u $
$=(\frac{(1-α)^{\alpha - 1}}{α^{\alpha - 2}}·\frac{(p_1)^{\alpha - 2}}{(p_2)^{\alpha - 1}})·p_1·(1 + (\frac{1-α}{α}·\frac{1}{1} ) )·u  $
$=(\frac{(1-α)^{\alpha - 1}}{α^{\alpha - 2}}·\frac{(p_1)^{\alpha - 1}}{(p_2)^{\alpha - 1}})·(\frac{\alpha}{\alpha} + (\frac{1-α}{α}) )·u  $
$=(\frac{(1-α)^{\alpha - 1}}{α^{\alpha - 2}}·\frac{(p_1)^{\alpha - 1}}{(p_2)^{\alpha - 1}})·(\frac{1}{α})·u  $
$=(\frac{(1-α)^{\alpha - 1}}{α^{\alpha - 1}}·\frac{(p_1)^{\alpha - 1}}{(p_2)^{\alpha - 1}})·u  $
$=(\frac{(1-α)}{α}·\frac{(p_1)}{(p_2)})^{\alpha - 1}·u  $
$=(\frac{α}{(1-α )}·\frac{p_2}{p_1})^{1-α}·u$
A: If you divide both sides by $u$ (so that factor just disappears), and multiply both sides by:
$$\frac{p_1}{\alpha}\left(\frac{1 - \alpha}{\alpha} \cdot \frac{p_1}{p_2}\right)^{1 - \alpha}$$
the identity to be proved becomes:
$$\frac{p_1}{\alpha} = p_1 + p_2 \cdot \frac{1 - \alpha}{\alpha} \cdot \frac{p_1}{p_2},$$
which is easily seen to be true.
