logarithmic inequality with different bases I have problem with following inequality $\log_{4}{5}+\log_{5}{6}+\log_{6}{7}+\log_{7}{8} \ge 4.4$
I tried to change all logarithms to base $10$ but it didn't work
 A: Rewrite the inequality as
$${1\over4}\left({\log5\over\log4}+{\log6\over\log5}+{\log7\over\log6}+{\log8\over\log7}\right)\gt{11\over10}$$
Now apply the AM-GM inequality:
$${1\over4}\left({\log5\over\log4}+{\log6\over\log5}+{\log7\over\log6}+{\log8\over\log7}\right)\ge\sqrt[4]{\log8\over\log4}=\sqrt[4]{3\over2}$$
It remains to observe that
$${3\over2}\gt\left({11\over10}\right)^4={14641\over10000}$$
A: You definitely have the right idea when it comes to a change of base. In general, let $a, b, x \in \mathbb{R}$ with $a,b \neq 1$; $a,b > 0$ and $x>0$. Then converting from base $a$ to base $b$ is done by the following:
$$ \log_a x = \frac{\log_b x}{\log_b a}$$ In your case, you want to convert each number to base 10 (so $b=10)$. In standard notation, $\log_{10}$ is written as just $\log$. Then your expression can be simplified like so:
$$\log_4 5  + \log_5 6 + \log_6 7 + \log_7 8 \geq 4.4 \\ \implies \frac{\log 5}{\log 4} + \frac{\log 6}{\log 5} + \frac{\log 7}{\log 6} + \frac{\log 8}{\log 7} \geq 4.4$$ and then see the answer from Barry Cipra.
A: $$\frac{\log5}{\log4}+\frac{\log6}{\log5}\cdots \approx 4.4289 $$
