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

I am not very good at math please explain it clearly

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  • $\begingroup$ What have you tried? Here's a good guide on how to ask a great question. $\endgroup$ – Toby Mak Sep 5 '17 at 10:31
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    $\begingroup$ What do you mean "compare"? I would think that you mean "which is larger?", but there are many ways to compare numbers. $\endgroup$ – Arthur Sep 5 '17 at 10:32
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    $\begingroup$ 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. $\endgroup$ – Arthur Sep 5 '17 at 10:47
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    $\begingroup$ I think it's clear that comparison in this context is talking about which is larger ... no need to be pedantic about it $\endgroup$ – Zubin Mukerjee Sep 5 '17 at 10:51
  • $\begingroup$ this question is answered here: math.stackexchange.com/questions/415500/… $\endgroup$ – lesnik Sep 6 '17 at 12:19
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Note: $$\log_2 3=\frac14 \log_2 81>\frac14\log_2 64=\frac64=\frac32,$$ $$\log_3 5=\frac14 \log_3 625<\frac14\log_3 729=\frac64=\frac32.$$

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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!

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$$\sqrt{3}\approx 1.73,\sqrt{2}\approx 1.42\\2^{1.5}=2\sqrt{2}< 2.84,\ 3^{1.5}=3\sqrt{3}>5.19\\\log_2 3>\log_22.84>\log_22\sqrt2= 1.5=\log_33\sqrt{3}> \log_35.19>\log_3 5$$

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