##log 12 ~~ 1.0792##

Since I remember (as is useful to remember):

##log 2 ~~ 0.30103##

##log 3 ~~ 0.47712125##

I can calculate:

##log 12 = log (2*2*3)##

##color(white)(log 12) = log 2 + log 2 + log 3##

##color(white)(log 12) ~~ 0.30103 + 0.30103 + 0.47712125##

##color(white)(log 12) ~~ 0.60206 + 0.47712125##

##color(white)(log 12) ~~ 1.07918125##

If you request ##log 12## on a calculator, you will get a similar approximation.

If we truncated this to ##4## decimal places then we would get:

##1.0791##

but the following digit is ##8 >= 5##, so in order to round the value to ##4## decimal places we need to round up the final digit ##1## to ##2## to get:

##log 12 ~~ 1.0792##

##color(white)()##**Footnote**

If you remember good approximations for ##log 2## and ##log 3## then you can calculate approximations to ##log n## for ##n in { 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20,… }##, i.e. any positive integer whose only prime factors are ##2##, ##3## or ##5##.

That seems to me to be good value for the sake of remembering a couple of numbers.