What can $\log_e(x) + \log_e(1+x) =0$ be written as? The equation $\log_e(x) + \log_e(1+x) =0$ can be written as:
a) $x^2+x-e=0$
b) $x^2+x-1=0$
c) $x^2+x+1=0$
d) $x^2+xe-e=0$
I tried differentiating both sides, then it becomes $\frac{1}{x}+\frac{1}{1+x}=0$, but I dont get any of the answers.

Recall that $$\log_e x + \log_e(x+1) = \log_e\Big( x(x+1)\Big) = \log_e(x^2+x)$$ and that $$ \text{if } \log_e(x^2+x) = 0 \text{ then } x^2+x = e^0. $$  
