How to solve $x^3 = x^{x^2-2x}$? Well i guess this is somehow pretty easy but there is something i don't understand.
I know that if $\;x>0\;$ then I can compare the exponents: $\; 3=x^2-2x$ , and from here I get that $\;x=3\; or \;x=-1\;$, but because $\;x>0\;$ that leaves me only with $\;x=3$ .
Second thing is that if both bases from both sides are equal to $1$ then its another solution, and therefore $\;x=1\;$ is another solution.
For conclusion we get that the solutions are: $\;x=3\; or \;x=1$ . 
Now my question is why $\;x=-1\;$ is also a solution? How do i get to this solution? am I suppose to just try place it and check because I somehow got it for $\;x>0$ ? are there any steps I can follow for solving this kind of equations?
Another question is how do you call this kind of equation? when I was looking for "exponential equations" I could only find ones with numbers in the bases .
Thanks!
 A: One approach. Note that $0$ is not a solution, if you follow the convention that $0^0=1$. So in the following, we may take the logarithm of $|x|$.
$$\begin{align}
x^3&=x^{x^2-2x}\\
\implies \left\lvert x^3\right\rvert&=\left\lvert x^{x^2-2x}\right\rvert\\
\implies |x|^3&=|x|^{x^2-2x}\\
\implies 3\ln|x|&=\left(x^2-2x\right)\ln|x|\\
\implies 0&=\left(x^2-2x-3\right)\ln|x|\\
\implies 0&=(x-3)(x+1)\ln|x|\\
\end{align}$$
So either $x=3$, $x=-1$, or $\ln|x|=0$. This last condition implies either $x=1$ or $x=-1$.
So (since this used one-way implications) there are three candidate solutions: $\{3,-1,1\}$. Then you can verify directly that each one is indeed a solution.
In the particular case of $-1$:  $$(-1)^3\stackrel{?}{=}(-1)^{(-1)^2-2(-1)}$$ is just true.
A: $x^a=x^b\implies a=b$ for all $x$,whether $x>0$ or $x<0$. You got $3=x^2-2x$, and solved this quadratic equation to get $x\in\{3,-1\}$. So $-1$ is indeed a solution. It depends on the equation as to which methods will you follow, there is no general algorithm. But yes, keep using exponent rules, and you will eventually lead to a solution.
A: look on rh(n)
and get x^2-2*x=3 (mod n)
Because rh(2)=-1 then we get 
(-1)^2+2=3=3 (mod 2)
and thats why it true 
