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

##### 1 Answer

Aug 19, 2017

#### Answer:

#### Explanation:

Suppose we know a suitable approximation for

#ln 10 ~~ 2.302585#

There is a series for

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

So putting

#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

#log(0.9863) ~~ -0.005991#