A) \[{{\log }_{e}}\left( \frac{m}{n} \right)\]
B) \[{{\log }_{e}}\left( \frac{n}{m} \right)\]
C) \[{{\log }_{e}}\left( \frac{m-n}{m+n} \right)\]
D) \[\frac{1}{2}{{\log }_{e}}\left( \frac{m}{n} \right)\]
Correct Answer: D
Solution :
\[\frac{m-n}{m+n}+\frac{1}{3}{{\left( \frac{m-n}{m+n} \right)}^{3}}+....\] = \[\frac{1}{2}{{\log }_{e}}\left( \frac{1+\frac{m-n}{m+n}}{1-\frac{m-n}{m+n}} \right)=\frac{1}{2}{{\log }_{e}}\frac{2m}{2n}=\frac{1}{2}{{\log }_{e}}\left( \frac{m}{n} \right)\].
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