Find the sum of all values of $x$ such that $\log_{3^2}(x-1)^2 = -1.$

If \(\log_9{(x-1)^2}=-1\) then \((x-1)^2=9^{-1}=\frac{1}{9}\)  so  \(x-1=\pm\frac{1}{3}\)

Can you take it from here?

Sorry but no

x - 1 = - 1/3

x = - 1/3 +1

x = 2/3

x - 1 =1/3

x =1/3 + 1

x = 4/3