Complex Numbers And Quadratic Equations question 748
Question: The equation $ {\log_{e}}x+{\log_{e}}(1+x)=0 $ can be written as [Kurukshetra CEE 1998; MP PET 1989]
Options:
A) $ x^{2}+x-e=0 $
B) $ x^{2}+x-1=0 $
C) $ x^{2}+x+1=0 $
D) $ x^{2}+xe-e=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ {\log_{e}}x+{\log_{e}}(1+x)=0 $
Þ $ {\log_{e}}(1+x)={\log_{e}}( \frac{1}{x} ) $
Þ $ x(x+1)=1\Rightarrow x^{2}+x-1=0 $
BETA
