<s>[INST] How to solve the equation for $x$ where $x^{(x-1)^2} = 2x +1 $ How to solve for $x$ in the following equation
$$ x^{(x-1)^2} = 2x +1$$
Guessing $(x-1)^2=2$ is a pretty good guess and it works giving $x = \sqrt{2} +1 $
But I want a non-guess solution. I tried thinking in terms of logarithm and binomial expansion but couldn't go far.
 [/INST] I'm assuming $x$ is non negative, since otherwise $x^{(x-1)^2}$ might not be well defined over the reals.
If $x<1$, then $x^{(x-1)^2}<1$, but $2x+1\ge 1$, so that can't yield a solution. Now suppose $x\ge 1$.
Let $y = (x-1)^2$, so $x^y = 2x+1 =$ $x^2+2-y$, therefore
$$x^y - x^2 = 2-y$$
Because $x \ge 1$, we have $x^y \le x^2$ if $y<2$ and $x^y \ge x^2$ if $y>2$, i.e.,

*

*If $y<2$, $x^y - x^2$ is non positive but $2-y$ is positive.

*If $y>2$, $x^y - x^2$ is non negative but $2-y$ is negative.

Hence $y = 2$ is the only candidate for a solution.
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