TITLE: $x^{x\sqrt{x}} = (x\sqrt{x})^x$: How can I correctly explain the existence of the root $x=1$?
QUESTION [5 upvotes]: This is a fairly well-known equation: $x^{x\sqrt{x}} = (x\sqrt{x})^x$
$$x^{x\sqrt{x}} = (x\sqrt{x})^x \iff x^{x^{3/2}}=(x^{3/2})^x \iff x^{x^{3/2}}=(x^{3x/2})$$
Take logarithms on both sides:
$$x^{3/2}\ln{x}=\frac{3x}{2}\ln{x}$$
From here it is quite easy to get one of the roots $x=\frac{9}{4}$ and $x=0$, if we divide both sides by $\ln{x}$.
However, why is $0$ not suitable, but $1$ is suitable?

REPLY [4 votes]: As soon as you used the logarithm function, you supposed that $x>0$. Indeed, $\ln$ is only defined on $]0,+\infty[$, which is why the solution $x=0$ should be excluded from your proof.
If you want your proof to be complete, you need to check if $0$ was a solution - here, we can see that it was (if you consider that $0^0$ is defined)
You were also wondering why the solution $x=1$ doesn't appear; it's because dividing by $\ln x$ means $\ln x \neq 0$, i.e. $x \neq 1$. Again, you need to check if $x=1$ is a solution afterwards - and it is!