$\log_3 4$ 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?
On
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 $$4>3^{1.2}.$$