# Solve $(x^2-5x+5)^{x^2-7x+12}= 1$.

What are all the possible values of $x$ if $(x^2-5x+5)^{x^2-7x+12}= 1$. I have already found three answers: 4, 3, and 1, however, apparently there are more possibilities. I don't know how to figure this out so it would be extremely appreciated if someone found the other possibilities and showed me how to do it.

• Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.)
– Blue
Aug 31 '18 at 1:18
• I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given. Aug 31 '18 at 1:29
• $x^2-7x+12 = (x-3)(x-4)$ Aug 31 '18 at 1:31
• For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed. Aug 31 '18 at 1:38
• No, wait. For $x^2-5x+5 < 0$ for $x=3$. Aug 31 '18 at 1:52

Hint:

$$(-1)^{k}=1$$ (if $$k$$ is an even integer).

In other words, you forgot the case that $$x^2-5x+5=-1.$$

Solving this, we get $$x^2-5x+6=0\rightarrow x=2,3.$$

We check that the exponent is even.

If $$x=3,$$ then $$(x-3)(x-4)=0.$$

If $$x=2,$$ then $$(-1)(-2)=2.$$

Either way, it is even, so both solutions are valid.

All solutions are $$1,2,3,4$$.

• So how would you use this to get the other possibilities? Sorry, I'm still quite confused. Aug 31 '18 at 1:33
• Solve for $x^2-5x+5=-1$ and check the exponent is even. Aug 31 '18 at 1:34
• In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution. Aug 31 '18 at 1:37
• @Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation? Aug 31 '18 at 1:59
• @JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $\log$ both side) why these are all solutions and there are no other. Can you explain : How we are sure there are no other solutions? Aug 31 '18 at 4:57