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?

$$\ln(x)+\ln(x-1)=\ln(2)\\
\implies \ln({x}({x-1}))=\ln(2)\\
\implies {x}({x-1})=2\\
\implies x^2-x-2=0\\
\implies x=\dfrac{1\pm3}{2}=-1,2$$
However, only $x=2$ satisifies the equation.
