jmc

Writing the equation in exponential form gives us $(x-1{)}^2=(3^2{)}^{-1}=3^{-2}=\frac{1}{9}$. Taking the square root of both sides of the equation $(x-1{)}^2=\frac{1}{9}$ gives $x-1=\pm \frac{1}{3}$. Solving $x-1=\pm \frac{1}{3}$ gives us $x=\frac{4}{3}\text{and}\frac{2}{3}.$ Therefore, our sum is $\frac{4}{3}+\frac{2}{3}=\boxed{2}.$