A problem that could use substitution or logs, not sure which works better

Question

This is one of those brain teaser problems on instagram, and it starts here:

$$x^{x^2-2x+1} = 2x + 1$$

And we want to solve for x. My first instinct was to try this

$$\ln(x^{x^2-2x+1}) = \ln(2x + 1)\\ (x^2-2x+1)\ln(x) = \ln(2x + 1)\\ (x-1)^2\ln(x) = \ln(2x + 1)\\ (x-1)^2 = \frac{\ln(2x + 1)}{\ln(x)}$$

But at that point I wasn't sure what rules apply; I know that the natural log $\ln(\frac{f(x)}{g(x)})=\ln(f(x))-\ln(g(x))$ but that doesn't work here. I could try

$$(x-1)^2 = \frac{\ln(2x + 1)}{\ln(x)}\\ (x-1)^2 = \frac{\ln(2(x + \frac{1}{2}))}{\ln(x)}=\frac{\ln(2)+\ln(x + \frac{1}{2})}{\ln(x)}$$

But htat seemed a dead end or over-complicated.

So another method was to say,

let $t= x-1$

So we get $$x^{x^2-2x+1} = x^{(x-1)^2} = x^{t^2}\\ x^{t^2} = 2x + 1\\$$

And we can rewrite $2x+1$ as $2(x-1+1)+1$ which turns it into $2(t+1)+1$

$$x^{t^2}= 2(t+1) + 1\\ (t+1)^{t^2} = 2(t+1)+1\\$$

But I feel a bit stuck here. I could try natural logs again

$$(t^2)\ln(t+1) = \ln(2(t+1)+1)=\ln(2t+3)$$

But I feel that I have gone in circles, and that there is some very simple piece of algebra I am missing. Anyhow, any guidance would be appreciated. I do get the sense that the form squared binomials take is important here; I know that I could just say that $t^2 = 2$ and that can work, but I am trying to get the last step that gets me to that conclusion; I feel there is a step missing.

Answer 1

Given that this is a "brain teaser", you might suspect that there's going to be at most quadratic equations to solve.
Next notice that if $x^2 - 2 x + 1 = 2$, $x^2 = 2 x + 1$. The solutions of $x^2 - 2 x + 1 = 2$ are $x = 1 \pm \sqrt{2}$.

Answer 2

Alpha doesn't find an algebraic solution, and neither did I, but the numeric answer it gets looks a lot like $1+\sqrt 2$ If we rewrite your equation as $$x^{(x-1)^2}=2x+1$$ we can plug into the left hand side to get $$(1+\sqrt 2)^{\sqrt 2 ^2}=(1+\sqrt 2)^2=3+2\sqrt 2=2(1+\sqrt 2)+1$$