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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.

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    $\begingroup$ Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.) $\endgroup$
    – Blue
    Aug 31 '18 at 1:18
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    $\begingroup$ I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given. $\endgroup$
    – vadim123
    Aug 31 '18 at 1:29
  • $\begingroup$ $x^2-7x+12 = (x-3)(x-4)$ $\endgroup$
    – amsmath
    Aug 31 '18 at 1:31
  • $\begingroup$ For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed. $\endgroup$
    – amsmath
    Aug 31 '18 at 1:38
  • $\begingroup$ No, wait. For $x^2-5x+5 < 0$ for $x=3$. $\endgroup$
    – amsmath
    Aug 31 '18 at 1:52
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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$.

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  • $\begingroup$ So how would you use this to get the other possibilities? Sorry, I'm still quite confused. $\endgroup$
    – Julie
    Aug 31 '18 at 1:33
  • $\begingroup$ Solve for $x^2-5x+5=-1$ and check the exponent is even. $\endgroup$
    – Jason Kim
    Aug 31 '18 at 1:34
  • $\begingroup$ 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. $\endgroup$
    – Jason Kim
    Aug 31 '18 at 1:37
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    $\begingroup$ @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? $\endgroup$
    – amsmath
    Aug 31 '18 at 1:59
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    $\begingroup$ @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? $\endgroup$ Aug 31 '18 at 4:57

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