Domain of $7^{\log_7(x^2-4x+5)}$ 
If $$7^{\log_7(x^2-4x+5)}=x-1$$ then $x$ may have values...

My attempt: $$x^2-4x+5=x-1$$ 
So, $$x^2-5x+6=0$$ 
So, $$x=2,3$$
To check the domain of log, $$x^2-4x+5>0$$ i.e., $$(x-2+i)(x-2-i)>0$$ That gives me, $x<2-i$ and $x>2+i$. Is this a valid way of writing domain here? If No, how to write the domain of $7^{\log_7(x^2-4x+5)}$? 
Also, if I put the value of $x$ as $2$ or $3$ in the given equation, it satisfies, but if I compare it with the inequalities $x<2-i$ or $x>2+i$, then I am not able to get a satisfactory answer.

Note that $$x^2-4x+5 = (x-2)^2 +1 >0 $$ for all x, so there is no problem with logarithm and we have
$$7^{log_7(x^2-4x+5)}=x^2-4x+5$$
Therefore $$ 7^{log_7(x^2-4x+5)}=x-1 \iff x^2-5x+6=0$$
Thus the  solutions are $x=2$ and $x=3$
