How do I solve $\vert x\vert^{x^2-2x} = 1$? I have the exponential equation $\vert x\vert^{x^2-2x} = 1$, but how do I solve it?
 A: I believe $x=0$ is a solution, because $x^x$ is continuous as $x$ approaches $0$. 
Consider $\lim_{x \to 0}x^x$.  Let $$f(x_n)=\bigg(\frac{1}{x}\bigg)^{\frac{1}{x}}$$such that
$$
y=\bigg(\frac{1}{x}\bigg)^{\frac{1}{x}}
$$Then
$$
ln(y)=\frac{1}{x}ln\bigg(\frac{1}{x}\bigg)=\frac{1}{x}(ln1-lnx)=\frac{1}{x}(0-lnx)=-\frac{lnx}{x}
$$Now, using L'Hopital's rule,
$$
\lim_{x\to\infty}\bigg(-\frac{lnx}{x}\bigg)=\lim_{x\to\infty}\bigg(-\frac{\frac{1}{x}}{1}\bigg)=\lim_{x\to\infty}\bigg(-\frac{1}{x}\bigg)=0
$$
Now,
$$
\lim_{x\to\infty}f(x_n)=\lim_{x\to\infty}y=\lim_{x\to\infty}e^{lny}=\lim_{x\to\infty}e^{-\frac{1}{x}}=e^0=1
$$Therefore,
$$
0^0=\lim_{x\to 0}x^x=\lim_{x\to 0}f(x)=\lim_{x\to\infty}f(x_n)=1
$$The solutions given by the graph provided by @CDspace are therefore correct:  {-1,0,1,2}
A: Either $x = \pm 1$ (so the value of the exponent doesn't really matter)
Or $x^2 -2x = 0$ but $x \neq 0$ i.e. $x = 2$ (any non-zero number to the zero-th power is one)
A: Hint:
note that your equation is equivalent to:
$$(x^2 - 2x) \, \log{|x|} = 0,$$
what can we say about a product of two factors if it is equal to zero?

Edit:
Suggested by @CDspace, here's a plot of $f(x) = |x|^{x^2-2x} - 1$ where we can visually see the solutions of $f(x) = 0$:

Pretty cool!
