Domain of $7^{\log_7(x^2-4x+5)}$

Question

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.

Answer 1

The tag says "complex numbers". In complex numbers you do not have any restriction that $x^2 - 4x +5$ must be positive. It could be negative or even non-real. But it can't be $0$.

so your only restriction is that $x \ne 2\pm i$.

Also there is no order in complex numbers. $x < 2-i$ does not make sense.

So if you are working with Complex numbers the domain is all $x$ so that $x^2 -4x +5 \ne 0$ or all complex except $2\pm i$. $2$ and $3$ are within that domain so you are safe.

If you are working with Real numbers the domain is all $x$ where $x^2 -4x + 5 = (x-2)^2 + 1 > 0$. This is all real numbers. $2$ and $3$ are within that domain so you are safe.

Answer 2

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 $x=3$