solving for $x^{\ln{(x)}}=1$

I think that I solved this equation, but there is still some dillema didi I did it right.

Can you please check my solution, and see did I make any mistakes, and maybe propose another solution.

$$x^{\ln{(x)}}=1\\ \ln(x^{\ln{(x)}})=\ln(1)\\ \text{I can write \ln(x^{\ln{(x)}}) as \ln(x)*\ln(x)\implies2\ln(x) and}\\ \text{\ln(1)\ as\ 0, so}\\ 2\ln(x)=0\\ \text{divide both sides with 2}\\ \ln(x)=0\\ \text{raising both sides to e}\\ e^{\ln(x)=e^0}\\ x=1\\$$

Thanks.

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why $ln(x)*ln(x) = 2 ln(x)$? –  JulianP Jan 4 at 15:13
Careful!! $\ln(x^2) = 2\ln(x)~~$ BUT $~~(\ln x)^2\neq \ln(x^2)$!! The final result is correct, but it comes from the fact that $$(\ln x)^2 = 0$$ which implies $\ln x = 0$ and so on, and not from $$2\ln x = 0$$ –  AndreasT Jan 4 at 15:13
$\ln(x) * \ln(x) = 2\ln(x) \implies x = e^2$ or $x = 1$, so it implies that it correct, but its not... –  JohnWO Jan 4 at 15:14

After what Andrea T wrote you are correctb,but you can do it more easily by reaching at the step $\ln^2(x)=0 \implies \ln x=0\implies x=1$ because $\ln x$ is one-to-one function.

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$$\ln x\cdot \ln x = (\ln x)^2\neq \ln(x^2) = 2\ln x$$

So what you have, instead, is $$(\ln x)^2 = \ln(1) = 0$$

Then, we do have that $(\ln x)^2 = 0 \implies \ln x = 0 \implies e^{\ln x} = e^0 = 1 \iff x = 1$

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Short and useful!+1 –  Sami Ben Romdhane Jan 4 at 15:22
@amWhy: I like direct and clean answers +1 –  Amzoti Jan 5 at 14:36

If you are allowing complex solutions, there are more.

\begin{align} &&x^{\ln(x)}&=1\\ \implies&& e^{\left(\ln(x)\right)^2}&=1\\ \implies&&\left(\ln(x)\right)^2 &=2\pi i n&\mbox{where n\in\mathbb{Z}}\\ \implies&& x&=e^{\pm\sqrt{2\pi i n}}\\ \implies&& x&=e^{\pm\sqrt{2\pi i n}} \end{align}

This can be further broken down into the form $e^A\left(\cos(B)+i\cos(C)\right)$ when you consider positive/negative values for $n$ and what $\pm\sqrt{\pm i}$ is.

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I think you should at least mention that $x\mapsto x^{\ln x}$ is not defined as a (single-valued) function for $x\in\Bbb C$, so these (for $x\neq1$) are only weak solutions, in the sense that one of the many possible values one could associate to the expression $x^{\ln x}$ equals $1$. –  Marc van Leeuwen Jan 5 at 8:08