How do you use a calculator to evaluate the expression #log12# to four decimal places?

1 Answer
George C.
Dec 25, 2016

#log 12 ~~ 1.0792#

Explanation:

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.

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