Question

$ {{\log }_{e}}x $ is equal to:

• A. $ (x-1)-\frac{{{(x-1)}^{2}}}{2}+\frac{{{(x-1)}^{3}}}{3}-....\infty $
• B. 0
• C. 1
• D. none of the above

Answer 1

Answer: (A)

The series of log, $(1+x) $ is
$ x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{2}}}{3}-\frac{{{x}^{4}}}{4}+.... $
Now, $ {{\log }_{e}}x={{\log }_{e}}(1+(x-1)) $
$ =(x-1)-\frac{{{(x-1)}^{2}}}{2}+\frac{{{(x-1)}^{3}}}{3}-.... $
Note: the series of $ \log \,(1+x) $ is defined only when $ -1 < x < 1. $