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Question

Question: \[1 + \frac{\log_{e}x}{1!} + \frac{(\log_{e}x)^{2}}{2!} + \frac{(\log_{e}x)^{3}}{3!} + .....\infty =...

1+logex1!+(logex)22!+(logex)33!+.....=1 + \frac{\log_{e}x}{1!} + \frac{(\log_{e}x)^{2}}{2!} + \frac{(\log_{e}x)^{3}}{3!} + .....\infty =

A

logex\log_{e}x

B

xx

C

x1x^{- 1}

D

loge(1+x)- \log_{e}(1 + x)

Answer

xx

Explanation

Solution

(0.5)(0.5)22+(0.5)33(0.5)44+....(0.5) - \frac{(0.5)^{2}}{2} + \frac{(0.5)^{3}}{3} - \frac{(0.5)^{4}}{4} + ....

loge32\log_{e}\frac{3}{2}.