How to prove $\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$ How could we prove that
$$ \frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$$
I have reduced it the form 
$$\frac{4^{\ln(4)/\ln(3/4)}}{3^{\ln(3)/\ln(3/4)}}$$
I am not sure what to do next to get snappy solution. Any ideas?
 A: This refers to the original question, which had the left hand side equal to $\frac{1}{2}$ instead of $\frac{1}{12}$.
They are not equal.
Your simplification is correct. Then we can rewrite the left hand side as
$$\left(\frac{4^{\ln(4)}}{3^{\ln(3)}}\right)^{1/\ln(3/4)}$$
so raising both sides of the equation to the $\ln(3/4)$ power, we get that the equation would be equivalent to
$$\frac{4^{\ln(4)}}{3^{\ln(3)}} \stackrel{?}{=} \frac{1}{2}^{\ln(3/4)}.$$
Rewriting $4^{\ln(4)}$ as $e^{(\ln 4)^2}$, $3^{\ln(3)}$ as $e^{(\ln(3))^2}$, and $\left(\frac{1}{2}\right)^{\ln(3/4)}$ as $e^{-\ln(2)\ln(3/4)}$, the equality would be equivalent to
$$\left(\ln 4\right)^2 - \left(\ln 3\right)^2 \stackrel{?}{=} -\ln(2)\ln\frac{3}{4}.$$
Now, $\ln(4) = 2\ln(2)$, and $\ln\frac{3}{4} = \ln 3 - 2\ln 2$. So the left hand side is equal to
$$4(\ln 2)^2 - (\ln 3)^2$$
while the right hand side is equal to
$$-\ln(2)(\ln 3 - 2\ln 2) = 2(\ln 2)^2 - (\ln 2)(\ln 3).$$
But $$4(\ln 2)^2 - (\ln 3)^2 \approx 0.714863$$
and
$$2(\ln 2)^2 - (\ln 2)(\ln 3) \approx 0.199406$$

As corrected, the right hand side now be, after the simplification
$$
\left(\frac{1}{12}\right)^{\ln(3/4)} = \exp\left(-\ln(12)\ln(3/4)\right).$$
The exponent can be simplified:
$$\begin{align*}
-\ln(12)\ln(3/4) &= -\left(\ln(3)+2\ln(2)\right)\left(\ln(3)-2\ln(2)\right)\\
&= \left(\ln(3)+2\ln(2)\right)\left(2\ln(2)-\ln(3)\right)\\
&= \left(2\ln(2)\right)^2 - \left(\ln 3\right)^2\\
&= 4(\ln 2)^2 - (\ln 3)^2.
\end{align*}$$
Since this is the same as the exponent of $e$ on the left hand side, we do indeed have
$$\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}.$$

There's nothing special about $3$ and $4$. Replacing them with arbitrary positive numbers $a$ and $b$ will lead to 
$$\exp\left((\ln(a))^2 - (\ln(b))^2\right) \stackrel{?}{=} \exp\left(-\ln(ab)(\ln(b/a)\right)$$
which of course holds, since
$$-\ln(ab)\ln(b/a) = (\ln a + \ln b)(\ln a - \ln b)$$
giving the equality you have in the comment:
$$\frac{a^{1/\log_a(b/a)}}{b^{1/\log_b(b/a)}} = \frac{1}{ab}.$$
A: This is another way, using the basics about logarithms.
You can use just the identity 
\begin{equation}
b^{\rm{log}_{\frac{b}{a}}b}=b\cdot a^{\rm{log}_{\frac{b}{a}}b}
\end{equation}
Note that $\frac{1}{\rm{log}_x{\frac{3}{4}}} =$ log$_\frac{3}{4} \ x$, you can use that to get
\begin{equation}
\frac{4^{\frac{1}{\rm{log}_4(3/4)}}}{3^{\frac{1}{\rm{log}_3(3/4)}}} = \frac{4^{\rm{log}_{3/4}4}}{3^{\rm{log}_{3/4}3}}
\end{equation}
Using the identity we have that
\begin{equation}
3^{\rm{log}_{\frac{3}{4}}3}=3\cdot 4^{\rm{log}_{\frac{3}{4}}3}
\end{equation}
replacing we get
\begin{equation}
\frac{4^{\rm{log}_{3/4}4}}{3^{\rm{log}_{3/4}3}}=\frac{4^{\rm{log}_{3/4}4}}{3\cdot 4^{\rm{log}_{\frac{3}{4}}3}}=\frac{4^{\rm{log}_{3/4}4-\rm{log}_{3/4}3}}{3}=\frac{4^{\rm{log}_{3/4}(4/3)}}{3}=\frac{4^{-1}}{3}=\frac{1}{12}
\end{equation}
A: $$\frac{4^{1/\log_4 3/4}}{3^{1/\log_3(3/4)}}=\frac{4^{\frac1{\log_43-1}}}{3^{\frac1{1-\log_34}}}=\frac{4^{\frac1{\log_43-1}}}{3^{\frac1{1-(\log_43)^{-1}}}}=\frac{4^{\frac1{\log_43-1}}}{3^{\frac{\log_43}{\log_43-1}}}=\left(\frac4{3^{\log_43}}\right)^{\frac1{\log_43-1}}$$
Taking the log base $4$, I get
$$\frac1{\log_43-1}\Big(1-(\log_43)^2\Big)=-(1+\log_43)\;.$$
Clearly $\log_4\frac1{12}=-\log_412=-(1+\log_43)$.
A: Equivalently, we want to prove that 
$$4^{1+\frac{1}{\log_4(3/4)}}=3^{\frac{1}{\log_3(3/4)}-1}.\tag{$1$}$$
We play a little with the left-hand side of $(1)$.
We have
$$4^{1+\frac{1}{\log_4(3/4)}}=4^{1+\frac{1}{\log_4 3-1}}=4^{\frac{\log_4 3}{\log_4 3-1}}=3^{\frac{1}{\log_4 3-1}}.$$
The right-hand side of $(1)$ can be written as
$$3^{\frac{\log_3 4}{1-\log_3 4}}.$$
We have expressed the left-hand side and the right-hand side as a power of $3$, and need to show that the exponents match. This is an easy consequence of the fact that $(\log_s t)(\log_t s)=1$.  
Remark: Of course there is nothing special about $3$ and $4$. Also, it would be more attractive to symmetrize, and write $4/3$ in some places, and $3/4$ in others.
