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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?

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    $\begingroup$ That's because the function $\ln(x)$ is only defined when $x > 0$. Moreover, $\ln(x)$ vanishes at $x = 1$, whence we conclude that $x = 1$ is also a solution. $\endgroup$
    – user0102
    Commented Jun 2, 2021 at 16:08
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    $\begingroup$ If you believe $0^0 = 1$, as I do, then $x=0$ is a solution to the original equation. (however, often $0^0$ is considered undefined and so it would not be a solution.) $\endgroup$
    – Jair Taylor
    Commented Jun 2, 2021 at 16:10

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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!

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  • $\begingroup$ Thank you! All the same, I don't teach the whole theory because of this, such simple mistakes appear =) $\endgroup$
    – QLimbo
    Commented Jun 2, 2021 at 16:17
  • $\begingroup$ If $0^0$ is undefined then $x=0$ is not a solution. $\endgroup$
    – Jair Taylor
    Commented Jun 2, 2021 at 16:19

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