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. 
 A: $\log_23$ 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 $\log_23<\log_36$
A: 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.
A: Here is mine...
$$\log_2 3=\frac{\log_2 243}5<\frac{\log_2 256}5=8/5$$
$$\log_3 2=\frac{\log_3 32}5>\frac{\log_3 27}5=3/5$$
$$\log_3 6=\log_3 2+\log_3 3>1+3/5=8/5>\log_2 3$$
A: Use:


*

*$$\log_2(3)=\frac{\ln(3)}{\ln(2)}$$

*$$\log_3(6)=\frac{\ln(6)}{\ln(3)}=\frac{\ln(2\cdot3)}{\ln(3)}=\frac{\ln(2)+\ln(3)}{\ln(3)}=1+\frac{\ln(2)}{\ln(3)}=1+\frac{1}{\log_2(3)}$$


So, you need to prove that:
$$\log_2(3)<\log_3(6)\space\space\space\Longleftrightarrow\space\space\space\log_2(3)<1+\frac{1}{\log_2(3)}$$
A: 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
$$\ln^2(3)<\ln(2\times 3)\ln 2=\ln(2)(\ln(2)+\ln(3)).$$
A: 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:
$$\begin{align}
5\log_23&=\log_2243\\
&\lt\log_2256\\
&=8\\
&=5+\log_327\\
&\lt5+\log_332\\
&=5\log_33+5\log_32\\
&=5\log_36
\end{align}$$
