<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] Here is mine...
$$\log_2 3=\frac{\log_2 243}5<\frac{\log_2 256}5=8/5$$
$$\log_3 2=\frac{\log_3 32}5>\frac{\log_3 27}5=3/5$$
$$\log_3 6=\log_3 2+\log_3 3>1+3/5=8/5>\log_2 3$$
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