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I tried to solve the problem below to get all the positive solutions: $$x^{x^2−3x} = x^2$$

By using $\ln$ on both sides, I get that one solution is $\displaystyle\frac{3 + \sqrt{17}}{2}$. But $1$ is also a solution that you can guess. How can I see it while solving the equation?

EDIT: I had a typo that showed i in front of $\displaystyle \frac{3 + \sqrt{17}}{2}$.

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    $\begingroup$ Is it more obvious if you rewrite the equation as $x^{x^2 - 3x - 2} = 1$? $\endgroup$ – Willie Wong May 10 at 5:26
  • $\begingroup$ $\mathrm i\bigl(3+\sqrt{17}\bigr)/2$ can’t be a “positive solution” because it isn’t real. It lies entirely on the imaginary axis, and $\Im\left\{\mathrm i\bigl(3+\sqrt{17}\bigr)/2\right\}$ is positive, if that’s what you mean. $\endgroup$ – let's have a breakdown May 10 at 7:54
  • $\begingroup$ Sorry, it's a typo. The i should not be there. I'll edit.- $\endgroup$ – Dovendyr May 10 at 13:17
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$$x^{x^2−3x} = x^2$$

Using $\ln$ on both sides, $$\implies (x^2-3x)\cdot\ln x = 2 \ln x$$

Here, you can divide by $\ln x$ if $\ln x \ne 0$, i.e., if $x \ne 1$. You have assumed that case while equating the powers.

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Take the $\ln$ of both sides.

You have $\ln x$ on each side of an equality and this immediately gives $x = 1$ as a solution. Factor out that $\ln x$ (for $x \neq 1$ solutions) and get a simple quadratic with imaginary solutions:

$$x = \frac{i}{2} \left(3 \pm \sqrt{17}\right)$$


I'd love to hear the justification of the downvote of my solution. Please post it as a comment, whoever you are.

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  • $\begingroup$ I'm m not the one who downvoted your answer, but I'll upvote it because it makes sense. $\endgroup$ – PranavGupta53535 May 10 at 5:55
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    $\begingroup$ Welcome to the club of victims of mysterious (and for sure anonymous) downvoters ! $\to +1$. $\endgroup$ – Claude Leibovici May 10 at 6:15
  • $\begingroup$ I voted up it as well to compensate. Thanks for replying :) $\endgroup$ – Dovendyr May 10 at 13:16

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