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

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)

• What if you expressed them in terms of a logarithm with the same base? Sep 25 '18 at 15:42
• @MatthewLeingang What base I need to choose? I tryed to take base 5 and have $\log_5 4 \cdot \log_5 6$ compare with 1. Sep 25 '18 at 15:47

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 $$\log_4{5} > \log_5{6}$$. $$\square$$

• @aid78's solution (+1) explains this better than I did. I just had a vague gut feeling that something like this "lemma" must be true, but [s]he has shown the intuition behind it: $\log_x{y} = \log_x(x(y/x)) = 1 + \log_x(y/x) > 1 + \log_x(v/u) \geqslant 1 + \log_u(v/u) = \log_u(u(v/u)) = \log_u(v)$. That's much nicer! Sep 25 '18 at 21:36

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.

$$\frac54>\frac65\land 4<5\implies\frac{\log\dfrac54}{\log 4}>\frac{\log\dfrac65}{\log 5}\implies\frac{\log5}{\log 4}>\frac{\log6}{\log 5}.$$