$\left( \frac{a - b}{a} \right) + \frac{1}{2} \left( \frac{a - b}{a} \right)^2 + \frac{1}{3} \left( \frac{a - b}{a} \right)^3 + \dots = $
- A$\log_e(a - b)$
- B$\log_e \left( \frac{a}{b} \right)$
- C$\log_e \left( \frac{b}{a} \right)$
- D$e^{\left( \frac{a - b}{a} \right)}$
Explore More
Similar Questions
$(0.5) - \frac{(0.5)^2}{2} + \frac{(0.5)^3}{3} - \frac{(0.5)^4}{4} + \dots$
Medium
View Solution$\log_e [(1 + x)^{1 + x} (1 - x)^{1 - x}] = $
Medium
View SolutionFor $|x| < 1$,the coefficient of $x^3$ in the expansion of $\log(1+x+x^2)$ in ascending powers of $x$ is (in $/3$)
The value of the infinite series $\log _4 2 - \log _8 2 + \log _{16} 2 - \dots \infty$ is:
Difficult
View SolutionThe expansion $\log_e(1 + x) = \sum\limits_{i = 1}^\infty \left[ \frac{(-1)^{i + 1}x^i}{i} \right]$ is defined for:
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo