log(x-1) + log(x+1) = 2log(x-2)

$$\log ((x-1)(x+1)) = 2 \log (x-2)$$ $$\log (x^2 - 1) = 2 \log (x-2)$$

$$x^2 - 1 = x^2 - 4x + 4$$

Subtract $x^2$ from both sides:

$$-1 = -4x + 4$$

Add $4x$ to both sides and add 1 to both sides:

$$4x = 4 + 1$$ $$4x = 5$$ $$x = \frac{5}{4}$$

For a logarithm $\log(A)$ to be defined, its argument $A$ must be positive ($A > 0$).
For $\log(x-1)$, we need $x-1 > 0 \implies x > 1$.
For $\log(x+1)$, we need $x+1 > 0 \implies x > -1$.
For $\log(x-2)$, we need $x-2 > 0 \implies x > 2$.

The most restrictive condition is $x > 2$.
Our calculated value is $x = \frac{5}{4} = 1.25$.
Since 1.25 is not greater than 2, this solution is extraneous and not valid for the original equation.

Therefore, there is no real solution for $x$.

$$\text{No solution}$$