What can $\log_e(x) + \log_e(1+x) =0$ be written as?

Question

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.

Answer 1

then you will get $$\ln(x)+\ln(1+x)=\ln(1)$$ or $$x(x+1)=1$$ can you finish?

Answer 2

For $x>0,$

$$\ln (ex)=\ln (e)+\ln (x)=1+\ln (x) $$

and $$-\ln (x)=\ln \Bigl(\frac {1}{x}\Bigr) $$

the equation will be

$$\ln (x+1)=\ln \Bigl(\frac 1x\Bigr) $$

or

$$x+1=\frac 1x $$ and the answer is $ x^2+x-1=0$

Answer 3

Since $\log(A) + \log(B) = \log(AB)$ $$ \begin{align} \log(x) + \log(1+x) &=0\\ \Rightarrow \log{\left( x(1+x)\right)} &=0 \\ \Rightarrow x(1+x) &= 1 \\ \Rightarrow x^2 + x -1 &=0 \end{align} $$

Answer 4

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