<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] From what you did, we want to prove that 
$$\frac{1}{1+\log_32}\lt \log_32,$$
i.e.
$$\log_32\gt \frac{\sqrt 5-1}{2},$$
i.e.
$$2\gt 3^{(\sqrt 5-1)/2},$$
i.e.
$$2^2\gt 3^{\sqrt 5-1},$$
i.e.
$$3\cdot 2^2\gt 3^{\sqrt 5}$$
It is sufficient to prove that
$$12\gt 3^{2.25},$$
i.e.
$$12^2\gt 3^4\sqrt 3,$$
i.e.
$$16\gt 9\sqrt 3$$
i.e.
$$16^2\gt 81\cdot 3$$
which holds.
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