how do you solve $x^x=x$

Question

I have tried to solve this by taking the natural log of both sides and got $x\ln(x)=\ln(x)$ and after I subtracted $\ln(x)$ from both sides I got $x\ln(x)-\ln(x)=0$ which using the distributive property becomes $(x-1)\ln(x)=0$ so either $x-1=0$ or $\ln(x)=0$ and in both cases $x=1.$ what I am confused about is that $(-1)^{-1}=(-1)$ and when I solved the equation I did not get this answer. Can somebody please tell me where I messed up and how to properly solve this equation?

Answer 1

When you take the natural logarithm of both sides you need to put the operand in an absolute value. Then your final equation would be $\ln|x|=0$ $\implies$ $x=1$ or $x=-1$.

Answer 2

When you take the log you are assuming that $x>0$.

$x^x=x\Longrightarrow x^{x-1}=1$

If $x-1=0$ (ie, $x=1$), then the equation is valid.

If $x$ is positive $\neq1$, then $x=1^{1/(x-1)}=1$, contradiction.

If $x$ is negative, we might have $|x|=1^{1/(x-1)}=1$, ie, $x=-1$. In fact, this is a solution.

Solution: $x=\pm1$.

OBS.: consider $0^0\neq0$ once $0^0$ is indetermined.