# Comparing logarithms with different bases

$$\log_3 4$$ and $$\log_7 10$$: which of these two logarithms is greater?

I figured out that both are between $$1$$ and $$2$$, then between $$1$$ and $$1.5$$. And then $$\log_34$$ is greater than $$1.25$$, and $$\log_710$$ is smaller than $$1.25$$. However, that method doesn't work for every example, and I wonder if there's a easier way to solve this?

• What have you tried? Are you stuck somewhere? – Andrei Oct 23 '18 at 21:01
• I figured out that both are between 1 and 2, then between 1 and 1.5. And then $/log_3 4$ is greater than 1.25 and $/log_7 10$ is smaller than 1.25. Howewer that method doesn't work for every example and i wonder if there's a easier way to solve this? – Hubert Hanc Oct 23 '18 at 21:09
• On second thoughts, this is not a duplicate of the other question, because none of the answers given in that case, including my own, can be applied in this case! (Ditto this.) It is interesting to ask if there is some general result that can be applied in this case; or at least, some other argument than the one given by the questioner himself (which I take it is essentially that $4^4 > 3^5$ and $10^4 < 7^5$, so $\log_34 > 5/4 > \log_710$). I suppose I should withdraw my close vote, if that's possible. Meanwhile, I've upvoted the question. – Calum Gilhooley Oct 24 '18 at 3:20
• To reduce the likelihood of the question being closed, perhaps the questioner should incorporate the text of his comment, showing work done on the problem, into the question itself. – Calum Gilhooley Oct 24 '18 at 3:22
• By a really strained application of a method used for the other question, one could argue that $\log_34 = \log_{27}64 > \log_{49}100 = \log_710$, because $64/27 > 2\tfrac{1}{3} > 100/49$ and $27 < 49$. But only a madman would do it that way! (Sorry, I mean only a mad person would do it that way.) :) – Calum Gilhooley Oct 24 '18 at 4:22

Here is a proof using no numbers larger than $$128$$. We have $$81 < 125 < 128$$, i.e. $$3^4 < 5^3 < 2^7$$, therefore $$\log_35 > \tfrac{4}{3}$$ and $$\log_25 < \tfrac{7}{3}$$, therefore $$\log_25 < 1 + \log_35$$, therefore: $$\log_210 < 2 + \log_35 = \log_345 < \log_349 = 2\log_37,$$ therefore $$\log_410 < \log_37$$, therefore $$\log_34 > \log_710$$. $$\square$$
The last step uses the general proposition that $$\log_ab > \log_cd$$ if and only if $$\log_bd < \log_ac$$, which can be proved by rewriting all the logarithms in terms of logarithms to a single base (e.g. $$\log_ab = \ln b/\ln a$$, etc.).
We'll show that $$\log_34>\log_710$$ or $$4>3^{\log_710},$$ which is true because $$\log_710<1.2$$ and $$4>3^{1.2}.$$