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

1 Answer
Aug 19, 2017

Answer:

#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#