Is the power-exponential equation insoluble? If yes, why? If no, what's the solution? $$\bigr(|x|+(-2)^{x-1})^{x-2}=1$$
I am a grade-7 middle school(Don't mind this, you can answer even if I can't understand it. But I will understand it.) student study in Shanghai, China. We are learning about integer power. And then I met a question on exam: $$(x-2)^{x-1}=1$$ It is clear that when $x=1$, the equation always stands up except the base $x-2=0$. But we have just defined $x=1$, so $x-2=-1\neq0$. Therefore, one of the solutions is $x=1$.
Then, consider $x-2=1$, since $1^x=1(x\in\Re)$. So we work out another solution $x=3$
After that, consider $x-2=-1$. The equation also stands up if $x-1$ is an even number at the same time, and it is true because $x=1$. But this solution is the same as the first one.
Therefore, this equation has 2 solutions, $x=1$ and $x=3$. And I begin to think that if there is a power-exponential equation which has only 1 or 0 solutions. It is easy to throw two kind of solutions: The solutions which make the base equal to -1 and the ones can make the exponent equal to 0. When the exponent is an even number, and the base is the same formula as the exponent at the same time, there will be only one kind of solution. And after a lot of time calculating, I worked out a "insoluble" power-exponential equation, and that's the one at the beginning. Nevertheless, when I tried to solve it, I found it is out of my ability. Therefore, I'm looking forward for someone help me calculate it out, and I will be very thankful.