Q. $\log_{e}\frac{1 + 3x}{1 - 2x}$ is equal to
A) $- 5x - \frac{5x^{2}}{2} - \frac{35x^{3}}{3} - \ \ldots$
B) $- 5x + \frac{5x^{2}}{2} - \frac{35x^{3}}{3} + \ \ldots$
C) $5x - \frac{5x^{2}}{2} + \frac{35x^{3}}{3} - \ \ldots$
D) $5x + \frac{5x^{2}}{2} + \frac{35x^{3}}{3} + \ \ldots$
Explanation
(c)
$\log_{e}\frac{1 + 3x}{1 - 2x} = \log_{e}{(1 + 3x)} - \log_{e}{(1 - 2x)}$
$= \left\lbrack 3x - \frac{(3x)^{2}}{2} + \frac{(3x)^{3}}{3} - ... \right\rbrack + \left\lbrack 2x + \frac{(2x)^{2}}{2} + \frac{(2x)^{3}}{3} + ... \right\rbrack$
$= 5x - \frac{5x^{2}}{2} + \frac{35x^{3}}{3} - ...$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.