# How do you evaluate #log 2#?

##### 1 Answer

One (somewhat impractical) way is to raise to the

#### Explanation:

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

If we raise

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

So the first digit of

#1024 / (10^3) = 1.024#

To find the next decimal place, evaluate

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

So the next decimal place is:

Then:

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

So the next decimal place is:

Divide

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

So the next decimal place is

#1.99506308^10 ~~ 999.002#

is just a shade under

So