7^log _7≤ft(x^2-4 x+5)=x-1, x may have values:

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.