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)
$\endgroup$ 25 Answers
$\begingroup$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 $$\log_56={\log 6\over \log 5}<{\log 5\over \log 4}=\log _45$$
$\endgroup$ $\begingroup$$$f(x) = \log_x(x+1)$$ is a strictly decreasing function for $x>1$.
You can see this by finding $f'(x)$ and noticing that $f'(x)<0$ for all $x>1$.
$\endgroup$ $\begingroup$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 $\log_4{5} > \log_5{6}$. $\square$
$\endgroup$ 1 $\begingroup$I have find one more solution$$\log_4 5 > \log_5 6$$$$\log_4 (4+1) > \log_5 (5+1)$$$$\log_4 4\cdot(1+0.25) > \log_5 5\cdot(1+0.2)$$$$1+\log_4 (1+0.25) > 1+ \log_5 (1+0.2)$$$$\log_4 (1+0.25) > \log_5 (1+0.2)$$$$\log_4 (1+0.25) > \frac{\log_4 (1+0.2)}{\log_4 5}$$$$\log_4 (1+0.25) > \log_4 (1+0.2) > \frac{\log_4 (1+0.2)}{\log_4 5}$$Q.E.D.
$\endgroup$ $\begingroup$$$\frac54>\frac65\land 4<5\implies\frac{\log\dfrac54}{\log 4}>\frac{\log\dfrac65}{\log 5}\implies\frac{\log5}{\log 4}>\frac{\log6}{\log 5}.$$
$\endgroup$
