The base-10 logarithm of 2 is the number #x# such that #10^x = 2#.

You can calculate logarithms by hand using just multiplication (and dividing by powers of 10 - which is just digit shifting) and the fact that #log_10 (x^10) = 10*log_10 x#, though it's not very practical...

It goes something like this:

#2 < 10# so #log_10(2) < 1#

So the part of #log_10(2)# before the decimal place is # **0** #.

Calculate #2^10# by multiplying it by itself to get:

#2^10 = 1024#

Notice that #10^3 = 1000 < 1024 < 10000 = 10^4#

So #3 = log_10(10^3) < log_10(2^10) < log(10_4) = 4#

Now #log_10(2^10) = 10 * log_10(2)#

so by raising to the #10th# power, we have multiplied the logarithm by 10.

So the first digit of #log_10(2)# after the decimal point is # **3** #.

Then

#log_10(2^10) - 3 = log_10(2^10) - log_10(1000)#

#= log_10(2^10/1000) = log_10(1024/1000) = log_10(1.024)#

So having found the digit #3#, we next divide #1024# by #10^3# to get #1.024# and repeat the process to get the next digit:

#1.024^10 ~= 1.2676506#

This is still less than #10#, so the next digit is # **0** #.

#1.2676506^10 ~= 10.715086#

#10^1 = 10 <= 10.715086 < 100 = 10^2#, so the next digit is # **1** #.

Divide #10.715086# by #10^1# to get #1.0715086#

#1.0715086^10 ~= 1.99506298#

This is less than #10#, so the next digit is # **0** #.

#1.99506298^10 ~= 999.0014 ~= 1000 = 10^3#

I'll stop here at this approximation, giving a digit # **3** #.

Collecting our digits, we have #log_10(2) ~= 0.30103#