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. 

Hint
You can use the fact that 
$$\log_a(b)=\frac{\ln(b)}{\ln(a)}$$
so you can rewrite your inequation
$$\frac{\ln(3)}{\ln(2)}<\frac{\ln(6)}{\ln(3)}$$
if and only if
$$\ln^2(3)<\ln(2\times 3)\ln 2=\ln(2)(\ln(2)+\ln(3)).$$
