Using Lagrange for finding Marshallian Demand I want to find the marshallian demand function for the user function $u(x_1,x_2) = x_1^ax_2^{1-a}$ where $a \in (0,1)$.
This is what I have so far:
$$L = x_1^ax_2^{1-a} - \lambda(p_1x_1 + p_2x_2 - y)$$
part. derivation with respect to $x_1$:
$$\frac{\delta L}{\delta x_1} = ax_1^{a-1}x_2^{1-a} - \lambda p_1 = 0$$
part. derivation with respect to $x_2$:
$$\frac{\delta L}{\delta x_2} = x_1^a(1-a)x_2^{-a} - \lambda p_2 = 0$$
rearrange derivation 1:
$$ax_1^{a-1}x_2^{1-a} = \lambda p_1$$
solve for $\lambda$
$$\frac{ax_1^{a-1}x_2^{1-a}}{p_1} = \lambda$$
rearrange derivation 2:
$$x_1^a(1-a)x_2^{-a} = \lambda p_2 $$
solve for $\lambda$
$$\frac{x_1^a(1-a)x_2^{-a}}{p_2} = \lambda$$
equalizing both equations:
$$\frac{ax_1^{a-1}x_2^{1-a}}{p_1} = \frac{x_1^a(1-a)x_2^{-a}}{p_2}$$
Dividing by the denominator:
$$(ax_1^{a-1}x_2^{1-a})p_2 = (x_1^a(1-a)x_2^{-a})p_1$$
I'm not sure, whether I'm on the right path. My final equation looks a bit scary. I would need to solve for $x_1$ at first, then put the result into the budget constraint.
As you can see, it would be very time consuming.
Is there an easier solution?
 A: A clever way to solve this kind of problems (with a Cobb-Douglas function) is as follows:
$ax_1^{a-1}x_2^{1-a} - \lambda p_1 = 0$
$x_1^a(1-a)x_2^{-a} - \lambda p_2 = 0$
Bringing the terms involving $\lambda$ to the RHS:
$ax_1^{a-1}x_2^{1-a} = \lambda p_1 $
$x_1^a(1-a)x_2^{-a} = \lambda p_2 $
Dividing the first equation by the second equation:
$\frac{ax_1^{a-1}x_2^{1-a}}{x_1^a(1-a)x_2^{-a}}=\frac{\lambda p_1}{\lambda p_2}$
$\lambda $ can be cancelled. In addition you can cancel out $x_1^{\alpha}$ and $x_2^{-\alpha}$.
$\frac{ax_1^{-1}x_2^{1}}{(1-a)}=\frac{ p_1}{ p_2}$
$\frac{ax_2^{1}}{(1-a)x_1^{1}}=\frac{ p_1}{ p_2}$
The exponents are not needed anymore.
$\frac{ax_2^{}}{(1-a)x_1^{}}=\frac{ p_1}{ p_2}$
$\color{blue}{***}$
Solving for $x_2p_2$
$x_2p_2=\frac{p_1(1-\alpha)}{\alpha}x_1$
Inserting the term for $x_2p_2$  in the budget restriction:
$y=x_1p_1+p_2x_2$
$y=x_1p_1+\frac{p_1(1-\alpha)}{\alpha}x_1$
Factoring out $x_1$
$y=x_1\left( p_1+\frac{p_1(1-\alpha)}{\alpha} \right)\Rightarrow y=x_1p_1+\frac{x_1p_1}{\alpha} -x_1\frac{p_1\alpha}{\alpha} $
The third term is the negative of the first term:
$y=\frac{x_1p_1}{\alpha} \Rightarrow x_1=\alpha\frac{y}{p_1}  $
For a given y and a given $p_2$ the maschallian demand function is $x_1(p_1,\overline{p}_2,\overline{y})=\alpha\frac{\overline y}{p_1}  $ 
The same can be done for $x_2(p_2,\overline{p}_1,\overline{y})$. Repeat the similar steps after $\color{blue}{***}$.
A: You should try taking the logarithm of $u(x_1,x_2)$. This is allowed because logarithms are monotonic transformations and preserve the preference relation on the set of consumption bundle. Use this to rewrite your utility function as a new function $v(x_1,x_2)$ and solve the Lagrangian multiplier problem accordingly. 
