jmc

Putting our logarithm into exponential notation gives $(x-1{)}^2=10-2x$. Expanding, we have

$$\begin{array}{rl}{x}^2-2x+1& =10-2x\\ \Rightarrow x^2+1& =10\\ \Rightarrow x^2& =9\\ \Rightarrow x& =\pm 3.\end{array}$$However, $x=-3$ does not work, because then the base of the logarithm would be $x-1=-3-1=-4$, and the base of a logarithm can't be negative. So our only solution is $x=\boxed{3}$.