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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)

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  • $\begingroup$ What if you expressed them in terms of a logarithm with the same base? $\endgroup$ Sep 25 '18 at 15:42
  • $\begingroup$ @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. $\endgroup$
    – aid78
    Sep 25 '18 at 15:47
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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 $$\log_56={\log 6\over \log 5}<{\log 5\over \log 4}=\log _45$$

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

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

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

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  • $\begingroup$ @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! $\endgroup$ Sep 25 '18 at 21:36
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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.

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

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