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?

$x=-1$ 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. 
