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. 

$\log_23$ vs $\log_36$

Multiply each term by $5$:

$5\log_23$ vs $5\log_36$

Apply logarithm rules:

$\log_23^5$ vs $\log_36^5$

Simplify:

$\log_2243$ vs $\log_37776$

Conclude:

$\log_2243<\log_2256=8=\log_36561<\log_37776$

Hence $\log_23<\log_36$
