Solve a system of equations with $\ln$ exponents $$2^{\ln x}=3^{\ln y}$$,   
$$(3x)^{\ln 4}=(2y)^{\ln 5}$$
I have tried exponentiation, but I seem to go in circles.
 A: Guide:
Rather than exponentiation, try taking logarithm. 
For example the first equation would become $$\color{blue}{(\ln x)}\ln 2 = \color{green}{(\ln y )}( \ln 3)\tag{1}$$ 
Do the same thing for the second equation too, then you can solve simultaneous equation to obtain $\ln x$ and $\ln y$.
Edit:
For the second equation: 
$$(\ln 4) (\ln 3 + \ln x) = (\ln 5) (\ln 2 + \ln y)$$
Rearranging, we have
$$ (\ln 4)(\ln x) - (\ln 5) (\ln y) = (\ln 5) (\ln 2) - \ln 4 \ln 3$$
$$2(\ln 2)\color{blue}{(\ln x)} - (\ln 5)\color{green}{ \ln y} = (\ln 2)(\ln 5-2\ln 3)\tag{2}$$
A: As the log function is injective, you can take the log of both sides; that is,
$$2^{\ln{x}}=3^{\ln{y}} \iff \ln{2^{\ln{x}}}=\ln{3^{\ln{y}}} \iff \ln{x}\cdot \ln{2}=\ln{y}\cdot \ln{3}$$
NOTE that it is crucial that the log function is injective for the application of this method.
To solve, assume $x=e^z>0$ and $y=e^w>0$ thus
$$\ln{e^z}\cdot \ln{2}=\ln{e^w}\cdot \ln{3} \iff z\cdot \ln{e}\cdot \ln{2}=w \cdot \ln{e}\cdot \ln{3} \iff z \cdot \ln{2}=w\cdot \ln{3} \iff$$ $$\frac{z}{w}=\frac{\ln{3}}{\ln{2}}$$
Thus all the solutions are:
$$z=w\frac{\ln{3}}{\ln{2}}$$ or $$w=z\frac{\ln{2}}{\ln{3}}$$
and since $z=\ln{x}$  and $w=\ln{y}$:
$$z=\frac{\ln{3}}{\ln{2}}e^y \iff x=e^{\frac{\ln{3}}{\ln{2}}\ln{y}}$$ 
or
$$w=\frac{\ln{2}}{\ln{3}}e^x \iff y=e^{\frac{\ln{2}}{\ln{3}}\ln{x}}$$ 
here is a graph of the solution

For the second equation you can proceed in the same way and try to figure out the solution.
A: You can also directly solve as follow:
$$2^{\ln x}=3^{\ln y} \iff {\ln x}\cdot{\ln 2}={\ln y}\cdot{\ln 3}$$
$$(3x)^{\ln 4}=(2y)^{\ln 5}\iff {\ln 3x}\cdot{\ln 4}={\ln 2y}\cdot{\ln 5} \iff ({\ln x}+{\ln 3})\cdot{\ln 4}=({\ln y}+{\ln 2})\cdot{\ln 5}$$
that is:
$$$$\begin{cases}
{\ln x}\cdot{\ln 2}-{\ln y}\cdot{\ln 3}=0 \\
{\ln x}\cdot{\ln 4}-{\ln y}\cdot{\ln 5}={\ln 2}\cdot{\ln 5}-{\ln 3}\cdot{\ln 4}
\end{cases}$$$$
setting ${\ln x}=X$ and ${\ln y}=Y$ the folowing linear system is obtained:
$$\begin{cases}
{\ln 2}\cdot X-{\ln 3}\cdot Y=0 \\
{\ln 4}\cdot X-{\ln 5}\cdot Y={\ln 2}\cdot{\ln 5}-{\ln 3}\cdot{\ln 4}
\end{cases}$$
which holds to the following solutions
$$\begin{cases}
X= -{\ln 3}={\ln \frac13} \\
Y= -{\ln 2}={\ln \frac12}
\end{cases}$$
and finally
$$\begin{cases}
x=\frac13 \\
y=\frac12
\end{cases}$$
