log_e x - log_e (x - 1) = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) આપણી પાસે $\log_e x - \log_e (x - 1) = \log_e \left( \frac{x}{x - 1} \right)$ છે.આને $\log_e \left( \frac{1}{1 - \frac{1}{x}} \right) = -\log_e \l

Vedclass · Accepted Answer

$\frac{1}{x} + \frac{1}{2x^2} + \frac{1}{3x^3} + \dots \infty $

Vedclass · Answer

(B) આપણી પાસે $\log_e x - \log_e (x - 1) = \log_e \left( \frac{x}{x - 1} \right)$ છે.આને $\log_e \left( \frac{1}{1 - \frac{1}{x}} \right) = -\log_e \left( 1 - \frac{1}{x} \right)$ તરીકે લખી શકાય છે.લઘુગણકીય શ્રેણીના વિસ્તરણ $-\log_e (1 - y) = y + \frac{y^2}{2} + \frac{y^3}{3} + \dots$ નો ઉપયોગ કરતા,જ્યાં $y = \frac{1}{x}$,આપણને મળે છે:$-\log_e \left( 1 - \frac{1}{x} \right) = \frac{1}{x} + \frac{1}{2x^2} + \frac{1}{3x^3} + \dots \infty$.

$\log_e x - \log_e (x - 1) = $

Similar Questions

જો $|a| < 1$ અને $b = \sum_{k=1}^{\infty} \frac{a^k}{k}$ હોય,તો $a$ ની કિંમત શું થાય?

$(0.5) - \frac{(0.5)^2}{2} + \frac{(0.5)^3}{3} - \frac{(0.5)^4}{4} + \dots$

$\frac{1}{n^2} + \frac{1}{2n^4} + \frac{1}{3n^6} + \dots \infty = $

$\frac{x - 1}{x + 1} + \frac{1}{2} \cdot \frac{x^2 - 1}{(x + 1)^2} + \frac{1}{3} \cdot \frac{x^3 - 1}{(x + 1)^3} + \dots \infty = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module