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.