### Instruction: Find the sum of all values of $x$ such that $\log_{3^2}(x-1)^2 = -1.$.### Response: This problem involves logarithms, so I need to remember the definition and properties of logarithms.
The definition of a logarithm is that $\log_b a = c$ means $b^c = a$.
So, in this problem, $\log_{3^2}(x-1)^2 = -1$ means $(3^2)^{-1} = (x-1)^2$.
I can simplify this by writing $3^2$ as $9$ and $9^{-1}$ as $\frac{1}{9}$.
Then I have $\frac{1}{9} = (x-1)^2$.
To solve for $x$, I need to take the square root of both sides.
But I have to be careful, because the square root can be positive or negative.
So I get $\pm \sqrt{\frac{1}{9}} = x-1$.
I can simplify the square root of $\frac{1}{9}$ by writing it as $\frac{1}{3}$.
Then I get $\pm \frac{1}{3} = x-1$.
To isolate $x$, I need to add $1$ to both sides.
So I get $x = 1 \pm \frac{1}{3}$.
This means there are two possible values of $x$: $1 + \frac{1}{3}$ and $1 - \frac{1}{3}$.
I can write these values as fractions with a common denominator of $3$.
Then I get $x = \frac{4}{3}$ or $x = \frac{2}{3}$.
These are the only two values of $x$ that satisfy the given equation.