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I found this problem interesting, namely we are given three values: $$\log_{1/3}{27}, \log_{1/5}{4}, \log_{1/2}{5}$$

We want to sort those values without using calculator. So I decided to work on them a little bit before comparing anything. Here is what I got:

First of all: $\log_{1/3}{27} = \log_{1/3}{3^3} = \log_{1/3}{(\frac{1}{3})^{-3}} = -3\\ \log_{1/5}{4} = \log_{1/5}{2^2} = 2 \cdot \log_{1/5}{2} = 2 \cdot \log_{5^{-1}}{2} = -2 \cdot \log_{5}{2}$

Since $\log_{5}{2} < 1, \text{then: } -2\cdot \log_{5}{2} > -2 $

And the third value: $\log_{1/2}{5} = \log_{2^{-1}}{5} = -\log_{2}{5}$

Since $2 < \log_{2}{5} < 3, \text{then: } -2 > -\log_{2}{5} > -3 $

This shows us that the first values is the smallest, then the third one and finally the second one: $$\log_{1/3}{27}, \log_{1/2}{5}, \log_{1/5}{4}$$

Is my approach correct or have I made some mistakes, can we get more accurate values for those logarithms without using calculator.

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  • $\begingroup$ Quite correct, though you can make it simpler. $\endgroup$ – Yves Daoust Sep 18 '18 at 6:37
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As an alternative and to check recall that

$$\log_b a=\frac{\log_c a}{\log_c b}$$

that is

$$\log_{\frac13}27=\frac{\log_2 27}{\log_2\frac13}=-\frac{\log_2 3^3}{\log_2 3}=-3$$

$$\log_{\frac15}4=\frac{\log_2 4}{\log_2 \frac15}=-\frac{\log_2 2^2}{\log_2 5}=-\frac{2}{\log_2 5}$$

$$\log_{\frac12}5=\frac{\log_2 5}{\log_2 \frac12}=-\frac{\log 5}{\log_2 2}=-\log_2 5$$

therefore as you stated

$$\log_{\frac13}27<\log_{\frac12}5<\log_{\frac15}4$$

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For clarity, we take the inverses of the bases, which will reverse the order ($\log_{1/x}y=-\log_x y$).

We have

$$\log_3{27}=3,$$

$$\log_54<1,$$ because $4<5^1$ and $$1<\log_25<3$$

because $2^1<5<2^3$.

The rest is yours.


Short answer:

$$\log_54<1<\log_25<3=\log_327$$

and

$$\log_{1/5}4>\log_{1/2}5>\log_{1/3}27.$$

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  • $\begingroup$ You can tighten the bounds on $ log_2 5 $. In fact $2^2 < 5 < 2^3$ $\endgroup$ – Martin Bonner supports Monica Sep 18 '18 at 12:09
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    $\begingroup$ Ah. Now I understand. You can tighten the bounds - but you don't need to $\endgroup$ – Martin Bonner supports Monica Sep 18 '18 at 12:12
  • $\begingroup$ @MartinBonner: exactly. $\endgroup$ – Yves Daoust Sep 18 '18 at 13:26

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