Sort those 3 logarithmic values without using calculator 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.

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