<s>[INST] 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?
 [/INST] 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}
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