Looking for alternative ways to solve $3^{2^x}=2^{3^x}$ 
If $\log2=m$ and $ \log3=n$, What is the answer of
$\large3^{2^x}=\large2^{3^x}$? $1)\log
 mn\quad\quad\quad\quad\quad\quad2)\frac{m-n}{\log m-\log
 n}\quad\quad\quad\quad\quad\quad3)\frac{\log \frac
 mn}{m-n}\quad\quad\quad\quad\quad\quad4)\frac{m+n}{\log mn}$

Here is my approach:
To solve the equation I took logarithm in base $3$ of both sides of the equation:
$$2^x=\log_32^{3^x}=3^x\log_32=3^x\times\frac{m}{n}$$
So we have $(\frac23)^x=\frac mn$. by taking logarithm in base $\frac23$ we have:
$$x=\log_{\large\frac23}(\frac mn)=\frac{\log(\frac mn)}{\log(\frac23)}=\frac{\log \frac mn}{m-n}$$

My questin: Is there other approach to solve this problem?
 A: You could take natural log. It will reduce the change of base steps
$\Rightarrow 2^x\log 3 = 3^x\log 2 \\\Rightarrow (\frac23)^x =\frac mn \\\Rightarrow x\log(\frac23) = x(m -n) = \log(\frac mn)\\
\Rightarrow \boxed{x = \frac{\log \frac mn}{m-n}}  $
A: Start  with $\large3^{2^x}=\large2^{3^x}$,
Take logarithm 1st time:
\begin{align}2^x\ln 3=3^x\ln2,  \frac{\ln 3}{\ln 2}= (\frac{3}{2})^x \end{align}
Take logarithm 2nd  time:
\begin{align}\ln (\frac{\ln 3}{\ln 2})=x(\ln 3-\ln 2),x=\frac{\ln (\frac{\ln 3}{\ln 2})}{\ln 3-\ln 2}\end{align}
Replace $\ln 3$  and $\ln 2$ with m and n:
\begin{align} x=\frac{\ln (\frac{m}{n})}{m-n}\end{align}
A: Semi-alternative approach : defer taking logarithms to the end.
$\log 2 = m, ~\log 3 = n \implies e^m = 2, 
~e^n = 3 \implies $ 
$\displaystyle 2^x = (e^m)^x = e^{mx}, ~3^x = (e^n)^x = e^{nx}.$
Therefore,
$\displaystyle 3^\left({2^x}\right) = 3^{\left(e^{mx}\right)} 
= \left(e^{nx}\right)^{\left(e^{mx}\right)}
= e^{(nx) \times \left(e^{mx}\right)}.$
Similarly,
$\displaystyle 2^\left({3^x}\right) = 2^{\left(e^{nx}\right)} 
= \left(e^{mx}\right)^{\left(e^{nx}\right)}
= e^{(mx) \times \left(e^{nx}\right)}.$
Therefore,
$$ e^{(nx) \times \left(e^{mx}\right)}
= e^{(mx) \times \left(e^{nx}\right)}.$$
Now, you can take logarithms of each side:
$\displaystyle (nx) \times \left(e^{mx}\right) 
= (mx) \times \left(e^{nx}\right) \implies $
$\displaystyle \frac{m}{n} = \frac{mx}{nx}
 = \frac{e^{mx}}{e^{nx}}.$
Taking logarithms again,
$\displaystyle \log{\frac{m}{n}} = mx - nx = (m-n)x \implies 
\frac{\log{\frac{m}{n}}}{m - n} = x.$
