How do you evaluate #log 2#?

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Answer 1 · 2015-10-24T09:40:10

One (somewhat impractical) way is to raise to the #10#th power repeatedly and compare with powers of #10# as follows:

#2 < 10^1# so #log(2) < 1# and the portion of #log(2)# before the decimal point is: #color(red)(0)#

If we raise #2# to the #10#th power, then the #log# of the resulting number will be #10 log(2)#, so

#10^3 = 1000 < 2^10 = 1024 < 10^4 = 10000#

So the first digit of #log(2)# after the decimal point is: #color(red)(3)#

#1024 / (10^3) = 1.024#

To find the next decimal place, evaluate #1.024^10# and compare it to powers of #10# to find:

#10^0 < 1.024^10 ~~ 1.2676506 < 10^1#

So the next decimal place is: #color(red)(0)#

Then:

#10^1 < 1.2676506^10 ~~ 10.71508605 < 10^2#

So the next decimal place is: #color(red)(1)#

Divide #10.71508605# by #10^1# then find:

#10^0 < 1.071508605^10 ~~ 1.99506308 < 10^1#

So the next decimal place is #color(red)(0)#

#1.99506308^10 ~~ 999.002#

is just a shade under #10^3#, so a good approximation for the next digit is: #color(red)(3)#

So #log(2) ~~ 0.30103#