Question

$\frac{x-y}{x}+\frac{1}{2}{{\left( \frac{x-y}{x} \right)}^{2}}+\frac{1}{3}{{\left( \frac{x-y}{x} \right)}^{3}}+....$ =

• A. ${{\log }_{e}}\,(x-y)$
• B. ${{\log }_{e}}\,(x+y)$
• C. ${{\log }_{e}}\,\left( \frac{x}{y} \right)$
• D. ${{\log }_{e}}\,\,xy$

Answer 1

Answer: (C)

${{\log }_{e}}\,\left( \frac{x}{y} \right)$

Answer 2

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
${{\log }_{e}}\left( 1-\frac{x-y}{x} \right)=-\left[ \frac{x-y}{x}+\frac{1}{2}{{\left( \frac{x-y}{x} \right)}^{2}} \right.$
$\left. +\frac{1}{3}{{\left( \frac{x-y}{x} \right)}^{3}}+.... \right]$
$\Rightarrow$ $\frac{x-y}{x}+\frac{1}{2}{{\left( \frac{x-y}{x} \right)}^{2}}+\frac{1}{3}{{\left( \frac{x-y}{x} \right)}^{3}}+....$
$=-{{\log }_{e}}\left( \frac{y}{x} \right)$
$={{\log }_{e}}\frac{x}{y}$