<s>[INST] Prove: $\log_{2}{3} < \log_{3}{6}$ How should I prove that $\log_{2}{3} < \log_{3}{6}$?
I tried something like this: $2^{\log_{2}{3}}< 2^{\log_{3}{6}}$, $3<6^{\log_{3}{2}}$,  $\log_{6}{3}<\log_{6}{6^{\log_{3}{2}}}=\log_{3}{2}$, $\frac{1}{\log_{3}{6}}< \log_{3}{2}$. $\frac{1}{1+\log_{3}{2}}<\log_{3}{2}$, but still nothing. 
 [/INST] This is essentially the same answer as barak manos's and Djura Marinkov's, mostly just presented in a different fashion, as one long string of self-explanatory equalities and inequalities:
$$\begin{align}
5\log_23&=\log_2243\\
&\lt\log_2256\\
&=8\\
&=5+\log_327\\
&\lt5+\log_332\\
&=5\log_33+5\log_32\\
&=5\log_36
\end{align}$$
</s>