# how do you solve $x^x=x$

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?

• $\ln(x)$ is only defined when $x>0$ unless you start working with complex numbers, so when you take the natural log, you're doing it under the assumption that $x>0$ – Brenton Jun 27 '19 at 21:44

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 $$x=-1$$.

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.

• 1^(1/(x-1)) is undefined for x=1 because it becomes 1^(1/0) and that is one of the intermediate forms. – Yay Jun 27 '19 at 21:48
• Sorry, I did a mistake. Many thanks! – Quiet_waters Jun 27 '19 at 21:57
• In line 4 you still have x=1^(1/(x-1))=1 so x=1, but if x=1 then x^(1/(x-1)) is equal to 1^(1/0) – Yay Jun 27 '19 at 22:13
• Sorry again, I hope it's fixed. – Quiet_waters Jun 27 '19 at 22:18