How do you solve #log_10 2#?

One somewhat cumbersome way of calculating #log_10 2# goes as follows:

First note that #2 < 10^1#, so #log_10 2 < 1# and we can write #color(blue)(0.)# as the start of the logarithm.

If we raise #2# to the #10#th power, then the effect on the logarithm is to multiply it by #10#.

Let's see what happens:

#2^10 = 1024 >= 1000 = 10^3#

So the first digit after the decimal point is #color(blue)(3)#

Divide #1024 / 10^3# to get #1.024#

To find the next digit of the logarithm, calculate #1.024^10# and compare it with powers of #10#:

#1.024^10 ~~ 1.26765060022822940149#

This is still less than #10^1#, so the next digit of the logarithm is #color(blue)(0)#

Then:

#1.26765060022822940149^10 ~~ 10.71508607186267320891 >= 10^1#

So the next digit of the logarithm is #color(blue)(1)#

Divide #10.71508607186267320891# by #10^1# to get:
#1.071508607186267320891#

Then:

#1.071508607186267320891^10 ~~ 1.99506311688075838379#

This is still less than #10^1#, so the next digit of the logarithm is #color(blue)(0)#

Then:

#1.99506311688075838379^10 ~~ 999.00209301438450246726#

That is very close to #1000=10^3#, so a very good approximation is to stop here with a digit #color(blue)(3)#, bearing in mind that it would actually be very slightly less.

Putting our digits together #log_10 2 ~~ 0.30103#