Question

How can you calculate the value of `log(0.9863)` ?

Answer 1

`log(0.9863) ~~ -0.005991`

Explanation:

Suppose we know a suitable approximation for `ln 10`, say:

`ln 10 ~~ 2.302585`

There is a series for `ln(1+x)` like this:

`ln(1+x) = x-x^2/2+x^3/3-x^4/4+...`

So putting `x=0.9863-1 = -0.0137`, we have:

`ln(0.9863) = (-0.0137)-(-0.0137)^2/2+(-0.0137)^3/3-(-0.0137)^4/4+...`

`color(white)(ln(0.9863)) = -0.0137-0.00018769/2-0.000002571353/3-...`

`color(white)(ln(0.9863)) = -0.0137-0.000093845-0.000000857117bar(6)-...`

`color(white)(ln(0.9863)) ~~ -0.013794702`

Then by the change of base formula:

`log(0.9863) = ln(0.9863)/ln(10) ~~ -0.013794702/2.302585 ~~ -0.00599096`

Since the argument `0.9863` was only quoted to `4` significant digits, we should probably limit the precision of our answer and say:

`log(0.9863) ~~ -0.005991`