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?

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 $x=2$.
