Compare $\log_34 $ and $\log_45.$ Let $x=\log_34 , y=\log_45.$
Then 
$$
\frac{3^x}{4^y}=\frac{4}{5}<1  \text{                 but } \frac{3^{x-1}}{4^{y-1}}=\frac{16}{15}>1 
$$
May I deduce now  that $\log_34 >\log_45?$
 A: It is not clear to me how you derive $\log_3 4 > \log_4 5$ from your argument. Consider $f(t) = \frac{\ln(t+1)}{\ln(t)}$ for $t > e$. We have
$$
f'(t) = \frac{\ln(t)\frac{1}{t+1} - \ln(t+1)\frac{1}{t}}{(\ln(t))^2} = \frac{1}{(\ln(t))^2t(t+1)}\cdot (\ln(t)t - \ln(t+1)(t+1)) < 0
$$
Therefore,
$$
f(3) > f(4) \quad\Rightarrow\quad \frac{\ln(4)}{\ln(3)} = \log_34 > \frac{\ln(5)}{\ln(4)} = \log_45
$$

Another proof. 
Let $x = \log_45$; that is, $4^x = 5$ and $x > 1$. We have $$3^{\log_34} = 4$$ and $$3^{\log_45} = 3^x = 5\cdot(\frac{3}{4})^x < 5 \cdot \frac{3}{4} = \frac{15}{4} < 4$$
Therefore, $\log_3 4 > \log_4 5$.
A: $$\frac{d^2}{dx^2}\log\log(x)=-\frac{1}{x^2\log^2(x)}-\frac{1}{x^2\log(x)}$$ is clearly negative if $x>1$, hence $\log\log(x)$ is a concave function on $(1,+\infty)$.
In particular 
$$ \log(x-1)\log(x+1) < \log^2(x) $$
holds for every $x>2$, implying that
$$ \log_{4}(5)=\frac{\log(5)}{\log(4)}<\frac{\log(4)}{\log(3)}=\log_{3}(4).$$
A: By convexity of the function $\log\log x$,
$$\log\log\frac{3+5}2>\frac{\log\log 3+\log\log5}2.$$
Then
$$\log^24>\log3\log5$$ and
$$\frac{\log4}{\log3}>\frac{\log5}{\log4}.$$
A: One more way: consider
$$
f(x) = \log _{x} (x+1) =  \frac{\ln (x+1)}{\ln x}\\
$$
I you take the limit as $x \to \infty$, it is $1$. Since we are only interested in  $x \geq 3, f(3)$ is positive and greater than 1. Now take the derivative: it is 
$$
f'(x) = -\frac{1}{x \log x} \bigg(\frac{1}{x+1} + \frac{\log (1+\frac{1}{x})}{\log x} \bigg)
$$ 
$f'(x)$ is negative for $x \geq3$, so the function is monotone decreasing, hence $\log_{3}4 > \log_{4}5$. 
