Logarithmic Equations and solving for the variable The equation is $\ln{x}+\ln{(x-1)}=\ln{2}$ . I have worked it all the way through, and after factoring the $x^2-1x-2$ I got $x=2$, $x=-1$, but my question is: Can we have both solutions or couldn't we have the negatives?
 A: Using the property of logaritm $\ln a+\ln b =\ln (a\cdot b)$ we obtain
$$\ln x(x-1)=\ln 2 $$
Now, for the property $\ln u =\ln v \Rightarrow u=v$ the obtain that
$$x(x-1)=2 \Rightarrow x^2-x-2=0 $$
Then, we obtain $x=2$ and $x=-1$, but $x=-1$ is not solution of the equation because $\ln (-1)\not \in \mathbb{R}$ (the function $f(x)=\ln x$ is defiined when $x>0$).
Conclusion: the equation $\ln x+\ln (x-1)=\ln 2$ have the solution $x=2$.
A: The easiest way is indeed to combine them:
$$
\ln \left[x(x-1)/2\right] = 0
$$
implying that $x(x-1) = 2$ or $x^2 - x - 2 = 0$, which indeed has exactly two solutions at $x = 2$ and $x= -1$.
But $x=-1$ cannot be a solution, since the original equation is only defined for $x \ge 1$, so the only solution is $x=2$.
A: $$\ln(x)+\ln(x-1)=\ln(2)\\
\implies \ln({x}({x-1}))=\ln(2)\\
\implies {x}({x-1})=2\\
\implies x^2-x-2=0\\
\implies x=\dfrac{1\pm3}{2}=-1,2$$
However, only $x=2$ satisifies the equation.
A: $x=-1$ is, in some sense, a solution of the equation $\ln x+\ln (x-1)=\ln2$. The problem is that $\ln(-1)$ and $\ln(-2)$ are complex numbers. But in complex analysis, logarithms are multivaluated. Namely:
$$\ln(-1)=(\pi+2k_1\pi)i$$
$$\ln(-2)=\ln2+(\pi+2k_2\pi)i$$
So $\ln(-1)+\ln(-2)=\ln 2+2(1+k_1+k_2)\pi i$. You can take certain branches of the logarithm to satisfy the equation, for example $k_1=-1$, $k_2=0$. But for other branches, it doesn't work.
Moreover, $x=-1$ is a solution of the equation $\ln(x^2-x)=\ln 2$. This equation is not the same as the other because the identity $\ln x+\ln(x-1)=\ln(x^2-x)$ only works for $x>0$. For $x<0$ we need those branches. 
