Evaluate: log _e(x + 1) - log _e(x - 1) = | vedclass

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(C) We know that $\log _e(x + 1) - \log _e(x - 1) = \log _e\left( \frac{x + 1}{x - 1} \right)$.By dividing the numerator and denominator by $x$,we get

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$2\left[ {\frac{1}{x} + \frac{1}{{3{x^3}}} + \frac{1}{{5{x^5}}} + \dots \infty } \right]$

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(C) We know that $\log _e(x + 1) - \log _e(x - 1) = \log _e\left( \frac{x + 1}{x - 1} \right)$.By dividing the numerator and denominator by $x$,we get:$\log _e\left( \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} \right) = \log _e\left( 1 + \frac{1}{x} \right) - \log _e\left( 1 - \frac{1}{x} \right)$.Using the logarithmic series expansion $\log _e(1 + y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \dots$ and $\log _e(1 - y) = -y - \frac{y^2}{2} - \frac{y^3}{3} - \dots$,where $y = \frac{1}{x}$:$\log _e\left( 1 + \frac{1}{x} \right) - \log _e\left( 1 - \frac{1}{x} \right) = 2\left( \frac{1}{x} + \frac{1}{3x^3} + \frac{1}{5x^5} + \dots \infty \right)$.

Evaluate: $\log _e(x + 1) - \log _e(x - 1) = $

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