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I have to solve the equation: $(x-2)^{x^2-6x+1}\leq 1 $.

If $x-2>1 $ i.e. $x> 3 $ we have $3< x \leq 3+2\sqrt{2}$.

If $0< x-2<1 $ i.e. $ 2< x<3$ we get no solutions.

If $x -2=1$ we get solution $x=3$ and if $x -2= -1$ we have $x=1$.

If $x=2$ we have $0^{-7} $ which is undefined.

Question: I wonder how this equation could be solved if $x<2$. I considered special case $x= 1$,but how could I solve it in general? Wolfram Alpha gives results $x = 3- \sqrt{n+8} , n \in \mathbb{Z}, n \geq -6$, but how can these solutions be obtained without program?

I found in some book that $a^x, a < 0$ is defined for $x = \frac{p}{q}, p \in \mathbb{Z}, q \in \mathbb{N}$, $q$ is odd. Can we apply that in some way and how. Please help,I would be greatly thankful.

User154
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  • If we solve $x^2-6x+1=n$ under $x\lt 2$, then we get $x = 3- \sqrt{n+8},n \geq -6$. – mathlove Nov 17 '20 at 14:31
  • Thank you, just one question: would it be complicated if we tried to solve $x^2-6x+1= \frac{p}{q}$? Would it lead us to some other solutions? Do we ever do that? I tried to find a solved example, but haven't found any. – User154 Nov 17 '20 at 14:37
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    Well, if you are looking for solutions like $x=-\frac13 \implies LHS=(-7/3)^{-8/9}$, then you run into multiple (complex) roots and need to select what real interpretation you need to decide on the inequality. Check say https://math.stackexchange.com/questions/317528/how-do-you-compute-negative-numbers-to-fractional-powers – Macavity Nov 17 '20 at 14:55
  • Maybe you can try to solve first the general equation $a^b\leq 1$ , and derive the various cases that have to be considered. – Floridus Floridi Nov 19 '20 at 08:30

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From the top answer in this question, it seems that only the second interpretation is useful here: the first one does not allow negative exponents for a negative base, and the third one yields complex results which cannot be compared with 1 in general. It seems that you are also referring to the second interpretation.

Hence in the case that the base is negative, $x^2-6x+1\in\mathbb Q$. Either from the rational roots theorem or from the fact that $\mathbb Z$ is integrally closed domain, $x^2-6x+1$ must be an integer.

Say $x^2-6x+1=n\in\mathbb Z$. Then $(x-3)^2=n+8$, which eventually yields the results given by Wolfram Alpha. Then use the inequality to bound $n$, and at last check what you get are in fact solutions.

Tesla Daybreak
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