How to see that $1$ is a solution of $x^{x^2−3x} = x^2$

Question

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 $\displaystyle \frac{3 + \sqrt{17}}{2}$.

Answer 1

$$x^{x^2−3x} = x^2$$

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.

Answer 2

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:

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

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