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Question

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

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

A

1x−12x2+13x3−.....∞\frac{1}{x} - \frac{1}{2x^{2}} + \frac{1}{3x^{3}} - .....\infty

B

1x+12x2+13x3+.....∞\frac{1}{x} + \frac{1}{2x^{2}} + \frac{1}{3x^{3}} + .....\infty

C

2(1x+13x3+15x5+...∞)2\left( \frac{1}{x} + \frac{1}{3x^{3}} + \frac{1}{5x^{5}} + ...\infty \right)

D

2(1x−13x3+15x5−...∞)2\left( \frac{1}{x} - \frac{1}{3x^{3}} + \frac{1}{5x^{5}} - ...\infty \right)

Answer

1x+12x2+13x3+.....∞\frac{1}{x} + \frac{1}{2x^{2}} + \frac{1}{3x^{3}} + .....\infty

Explanation

Solution

e4x−1e2x\frac{e^{4x} - 1}{e^{2x}}

x2x^{2}

12\frac{1}{2}