I need to compare $\log_4 5$ and $\log_5 6$.
I can estimate both numbers like $1.16$ and $1.11$. Then I took smallest fraction $\frac{8}{7}$ which is greater than $1.11$ and smaller than $1.16$ and proove two inequalities: $$\log_4 5 > \frac{8}{7}$$
$$\frac{7}{8}\log_4 5 > 1$$
$$\log_{4^8} 5^7 > 1$$
$$\log_{65536} 78125 > 1$$
and
$$\log_5 6 < \frac{8}{7}$$
$$\frac{7}{8}\log_5 6 < 1$$
$$\log_{5^8} 6^7 < 1$$
$$\log_{390625} 279936 < 1$$
thats why I have $\log_5 6 < \frac{8}{7} < \log_4 5$.
But for proving I need estimation both logarithms (without this estimation I cannot find the fraction for comparing).
Can you help me to find more clear solution (without graphs)
Best Answer
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 $$\log_56={\log 6\over \log 5}<{\log 5\over \log 4}=\log _45$$