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?

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    $\begingroup$ $\ln(x)$ 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 $x>0$ $\endgroup$ – 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.

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    $\begingroup$ 1^(1/(x-1)) is undefined for x=1 because it becomes 1^(1/0) and that is one of the intermediate forms. $\endgroup$ – Yay Jun 27 '19 at 21:48
  • $\begingroup$ Sorry, I did a mistake. Many thanks! $\endgroup$ – Quiet_waters Jun 27 '19 at 21:57
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    $\begingroup$ 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) $\endgroup$ – Yay Jun 27 '19 at 22:13
  • $\begingroup$ Sorry again, I hope it's fixed. $\endgroup$ – Quiet_waters Jun 27 '19 at 22:18

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