Sequence And Series Question 56
Question: $ {\log_{e}}x-{\log_{e}}(x-1)= $
Options:
A) $ \frac{1}{x}-\frac{1}{2x^{2}}+\frac{1}{3x^{3}}-…..\infty $
B) $ \frac{1}{x}+\frac{1}{2x^{2}}+\frac{1}{3x^{3}}+…..\infty $
C) $ 2,( \frac{1}{x}+\frac{1}{3x^{3}}+\frac{1}{5x^{5}}+…\infty ) $
D) $ 2,( \frac{1}{x}-\frac{1}{3x^{3}}+\frac{1}{5x^{5}}-…\infty ) $
Show Answer
Answer:
Correct Answer: B
Solution:
$ {\log_{e}}x-{\log_{e}}(x-1) $ $ ={\log_{e}}( \frac{x}{x-1} )={\log_{e}}( \frac{1}{1-\frac{1}{x}} )=-{\log_{e}}( 1-\frac{1}{x} ) $ $ =\frac{1}{x}+\frac{1}{2x^{2}}+\frac{1}{3x^{3}}+…. $
BETA
