<s>[INST] 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 $\log_34 >\log_45?$
 [/INST] 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, $\log_3 4 > \log_4 5$.
</s>