A) \[{{x}^{2}}+x-e=0\]
B) \[{{x}^{2}}+x-1=0\]
C) \[{{x}^{2}}+x+1=0\]
D) \[{{x}^{2}}+xe-e=0\]
Correct Answer: B
Solution :
\[{{\log }_{e}}x+{{\log }_{e}}(1+x)=0\]Þ \[{{\log }_{e}}(1+x)={{\log }_{e}}\left( \frac{1}{x} \right)\] Þ \[x(x+1)=1\,\,\Rightarrow \,\,{{x}^{2}}+x-1=0\]
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