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Question: \[\log_{e}(x + 1) - \log_{e}(x - 1) =\]...

log⁡e(x+1)−log⁡e(x−1)=\log_{e}(x + 1) - \log_{e}(x - 1) =

A

2[x+x33+x55+......∞]2\left\lbrack x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + ......\infty \right\rbrack

B

[x+x33+x55+......∞]\left\lbrack x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + ......\infty \right\rbrack

C

2[1x+13x3+15x5+...∞]2\left\lbrack \frac{1}{x} + \frac{1}{3x^{3}} + \frac{1}{5x^{5}} + ...\infty \right\rbrack

D

[1x+13x3+15x5+...∞]\left\lbrack \frac{1}{x} + \frac{1}{3x^{3}} + \frac{1}{5x^{5}} + ...\infty \right\rbrack

Answer

2[1x+13x3+15x5+...∞]2\left\lbrack \frac{1}{x} + \frac{1}{3x^{3}} + \frac{1}{5x^{5}} + ...\infty \right\rbrack

Explanation

Solution

1log⁡e(1−y)\frac{1}{\log_{e}(1 - y)}

log⁡e11−y\log_{e}\frac{1}{1 - y}.