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.
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.