# Calculus!!!

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Which of the following series can be used to compute ln(.8)?
c) ln x in powers of (x-1)
d) ln(x-1) in powers of (x-1)
e) none of the proceding

I am not certain any of our volunteers have expandied series of log functions in the recent past, and would likely give you misleading info. Perhaps one of the non regular math profs will drop in and help you.

The Taylor Series says:
ln(x)== ln(a) + (x-a) / a - (x-a)2 / 2a2 + (x-a)3 / 3a3 - (x-a)4 / 4a4 + ...

where a is a value for which you know ln(a)
your x=.8, then I would choose a=1
and
ln(.8)=ln(1)+(-.2)/1+(-.2)^2/2+(-.2)^3/3...
=0+a series of negative terms.

I did this for 5 terms and got -.2233866
with the real value of ln(.8)=-.22314..

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