$7^{\log _7\left(x^2-4 x+5\right)}=x-1$, x may have values:

2, 3 7 -2, -3 2, -3

2, 3

We have $7^{\log _7\left(x^2-4 x+5\right)}=x-1$ $\Rightarrow x^2-4 x+5=x-1\left(∵ a^{\log _a x}=x\right)$ ⇒ (x − 3)(x − 2) = 0 ⇒ x = 2, 3 Hence (1) is the correct answer.