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