Q. $\frac{x - y}{x} + \frac{1}{2}\left( \frac{x - y}{x} \right)^{2} + \frac{1}{3}\left( \frac{x - y}{x} \right)^{3} + ...$ is equal to

A) $\log_{e}{(x - y)}$
B) $\log_{e}{(x + y)}$
C) $\log_{e}\left( \frac{x}{y} \right)\$
D) $\log_{e}{xy}$
Explanation

(c)
Put $X = \frac{x - y}{x}$ in $\log_{e}{(1 - X)} = - \left\lbrack \frac{X}{1} + \frac{X^{2}}{2} + \frac{X^{3}}{3} + ... \right\rbrack$
We get
$\log_{e}\left( 1 - \frac{x - y}{x} \right) = - \left\lbrack \frac{x - y}{x} + \frac{1}{2}\left( \frac{x - y}{x} \right)^{2} + \frac{1}{3}\left( \frac{x - y}{x} \right)^{3} + ... \right\rbrack$
$\Rightarrow \frac{x - y}{x} + \frac{1}{2}\left( \frac{x - y}{x} \right)^{2} + \frac{1}{3}\left( \frac{x - y}{x} \right)^{3} + ... = - \log_{e}\left( \frac{y}{x} \right) = \log_{e}\frac{x}{y}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.