A) \[{{\log }_{e}}1\]
B) \[{{\log }_{e}}2\]
C) \[{{\log }_{e}}3\]
D) \[{{\log }_{e}}4\]
Correct Answer: C
Solution :
Since,\[\log (1+x)-\log (1-x)\] \[=2\left[ x+\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}+...\infty \right]\] Put \[x=\frac{1}{2}\]on both sides, we get \[\log \left( \frac{3}{2} \right)-\log \left( \frac{1}{2} \right)=2\left( \frac{1}{2}+\frac{1}{3}.\frac{1}{{{2}^{3}}}+\frac{1}{5}.\frac{1}{{{2}^{5}}}+...\infty \right)\] \[\Rightarrow \] \[\log 3=1+\frac{1}{3}.\frac{1}{4}+\frac{1}{5}.\frac{1}{{{4}^{2}}}+...\]
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