Sequence And Series Question 38
Question: $ {\log_{e}}(x+1)-{\log_{e}}(x-1)= $
Options:
A) $ 2,[ x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+……\infty ] $
B) $ ,[ x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+……\infty ] $
C) $ 2,[ \frac{1}{x}+\frac{1}{3x^{3}}+\frac{1}{5x^{5}}+…\infty ] $
D) $ ,[ \frac{1}{x}+\frac{1}{3x^{3}}+\frac{1}{5x^{5}}+…\infty ] $
Show Answer
Answer:
Correct Answer: C
Solution:
$ {\log_{e}}(x+1)-{\log_{e}}(x-1)={\log_{e}}\frac{x+1}{x-1} $ $ ={\log_{e}}( \frac{1+\frac{1}{x}}{1-\frac{1}{x}} )=2{ \frac{1}{x}+\frac{1}{3x^{3}}+\frac{1}{5x^{5}}+……. } $ .
BETA
