# How can I compare $\log_2 3$ and $\log_3 5$ without using a calculator [closed]

Compare $\log_2 3$ and $\log_3 5$ without using a calculator.

I am not very good at math please explain it clearly

• What have you tried? Here's a good guide on how to ask a great question. – Toby Mak Sep 5 '17 at 10:31
• What do you mean "compare"? I would think that you mean "which is larger?", but there are many ways to compare numbers. – Arthur Sep 5 '17 at 10:32
• For the upvoters: This might be an interesting problem, but it is, at the moment, a very poorly phrased question. If you want to show it recognition for being an interesting question, or you are interested in knowing the answer, consider favouriting instead. – Arthur Sep 5 '17 at 10:47
• I think it's clear that comparison in this context is talking about which is larger ... no need to be pedantic about it – Zubin Mukerjee Sep 5 '17 at 10:51
• this question is answered here: math.stackexchange.com/questions/415500/… – lesnik Sep 6 '17 at 12:19

Note: $$\log_2 3=\frac14 \log_2 81>\frac14\log_2 64=\frac64=\frac32,$$ $$\log_3 5=\frac14 \log_3 625<\frac14\log_3 729=\frac64=\frac32.$$
We'll prove that $$\log_35<\log_23$$ or $$5<3^{\log_23}$$ or $$25<3^{\log_29},$$ which is true because $$3^{\log_29}>3^{\log_28}=27>25.$$ Done!
$$\sqrt{3}\approx 1.73,\sqrt{2}\approx 1.42\\2^{1.5}=2\sqrt{2}< 2.84,\ 3^{1.5}=3\sqrt{3}>5.19\\\log_2 3>\log_22.84>\log_22\sqrt2= 1.5=\log_33\sqrt{3}> \log_35.19>\log_3 5$$