How to compare logarithms $\log_4 5$ and $\log_5 6$?

Use the Am-Gm inequality and the fact that $\log x$ is increasing:

$$\log 6\cdot \log 4< {(\log 6+\log 4)^2\over 4} ={\log^2 24\over 4} < {\log ^225\over 4 }= \log ^25$$

So $$\log_56={\log 6\over \log 5}<{\log 5\over \log 4}=\log _45$$

$$f(x) = \log_x(x+1)$$ is a strictly decreasing function for $x>1$.

You can see this by finding $f'(x)$ and noticing that $f'(x)<0$ for all $x>1$.

Lemma If $v \geqslant u \geqslant x > 1$ and $y/x > v/u$, then $\log_x{y} > \log_u{v}$.

Proof Let $\alpha = \log_x{y}$, and $\beta = \log_u{v} \geqslant 1$. Then $x^{\alpha-1} = y/x > v/u = u^{\beta-1} \geqslant x^{\beta-1}$, therefore $\alpha > \beta$. $\square$

We have $5/4 > 6/5$, so the lemma gives $\log_4{5} > \log_5{6}$. $\square$