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

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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

$\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 = $

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

શ્રેણી $x \log _e a + \frac{x^3}{3!} (\log _e a)^3 + \frac{x^5}{5!} (\log _e a)^5 + \dots$ નું મૂલ્ય શું છે?

$\frac{1}{2} - \frac{1}{2 \cdot 2^2} + \frac{1}{3 \cdot 2^3} - \frac{1}{4 \cdot 2^4} + \ldots$ ની કિંમત શોધો.

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