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

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

• Is it more obvious if you rewrite the equation as $x^{x^2 - 3x - 2} = 1$? – Willie Wong May 10 at 5:26
• $\mathrm i\bigl(3+\sqrt{17}\bigr)/2$ 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. – let's have a breakdown May 10 at 7:54
• Sorry, it's a typo. The i should not be there. I'll edit.- – Dovendyr May 10 at 13:17

$$x^{x^2−3x} = x^2$$

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.

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.

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