Solve $\log_2{\log_x{(x-3y)}}=-1, x\cdot y^{\log_x{y}}=y^{\frac{5}{2}}$ Problem is to solve this system of equation.
$$\log_2{\log_x{(x-3y)}}=-1$$
$$x\cdot y^{\log_x{y}}=y^{\frac{5}{2}}$$
\My attempt:
First equation can be written as $$\log_x{(x-3y)}=\frac{1}{2}$$.
If on the second equation I put $\log_x()$ I get this
$$\log_x{x\cdot y^{\log_x{y}}}=\log_x{y^\frac{5}{2}}$$
$$\log_x{y} + \log_x{y^{\log_x{y}}}=\frac{5}{2}\log_x{y}$$
$$\log_x{y} + \log_x{y}\cdot\log_x{y}=\frac{5}{2}\log_x{y}$$
$$t+t^2-\frac{5}{2}t=0$$
$$t^2 - \frac{3}{2}t=0$$
There are two solutions $t=0$ and $t=\frac{3}{2}$.
$$\log_x{y}=0$$
From that I can easily see that $y=1$.
But this information didn't help me to solve $\log_x{x-3}=\frac{1}{2}$ because I didn't know how to solve this.

You have some mistakes when you solve the second one. A $x$ is changed to a $y$. But the approach is correct. First of all $y$ is positive, which means $x$ has to be as well. Then
$$x\cdot y^{\log_x{y}}=y^{\frac{5}{2}}\Rightarrow \log_x \color{red}{x}+\log_x y\cdot \log_x y=\frac{5}{2}\log_x y$$
or 
$$1+t^2=\frac{5}{2}t$$
with $t = \log_x y$. Solving gives $t=2$ or $t=\frac{1}{2}$. 
If $t=2$, we have $y=x^2$, so substituting in the first equation:
$$\log_2{\log_x{(x-3x^2)}}=-1$$
or
$$\log_x(x-3x^2)=\frac{1}{2}$$
or $x-3x^2=\sqrt{x}$. However this is impossible, because from AM-GM:
$$3x^2+\sqrt{x}=3x^2+\frac{1}{2}\sqrt{x}+\frac{1}{2}\sqrt{x}\geq 3\sqrt[3]{\frac{3}{4}}x>x$$
which means no solution in this case. If $t=\frac{1}{2}$, we have $y=\sqrt{x}$, so:
$$\log_x(x-3\sqrt{x})=\frac{1}{2}$$
or $x-3\sqrt{x}=\sqrt{x}$, which gives the solution $x=16$ and $y=4$.
