TITLE: How do I solve $\vert x\vert^{x^2-2x} = 1$?
QUESTION [11 upvotes]: I have the exponential equation $\vert x\vert^{x^2-2x} = 1$, but how do I solve it?

REPLY [2 votes]: 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}