<s>[INST] equation with quadratic power I was thinking how to solve:


*

*If $x^{(x-1)^2}=2x+1$, find $x-\frac{1}{x}$

*Solve $x^{x-x^2+13} = x^2-12$
I noticed that in both problems, the linear part can be constructed in the quadratic exponent, I tried a few change of variable and nothing.
 [/INST] Here's a possible approach to the second problem:
Observe that $x-x^2+13=-(x^2-12)+x+1$, so let $a=x^2-12$.
Now, we wish to solve the equation $x^{-a+x+1}=a$. Note that $x=0$ is not a solution; this is by inspection.
It is easy to see that $x=a$ actually satisfies the preceding equation. Now, we argue that all other values of $x$ does not work:
$x>a \Rightarrow x-a >0 \Rightarrow x-a+1>1 \Rightarrow x^{x-a+1}>a^{x-a+1} >a \ \forall \ x$, if $x \in \mathbb{R^+}$.  So, equality does not hold. 
Likewise, $x<a \Rightarrow x-a <0 \Rightarrow x-a+1<1 \Rightarrow x^{x-a+1}<a^{x-a+1} <a \ \forall \ x$, if $x \in \mathbb{R^+}$, and $x-a+1 \geq 0$. Otherwise, if $x-a+1<0$, $ x^{x-a+1}>a^{x-a+1} >a$. So equality never happens.
Thus, given that $x=a$, we have $x^2-12=x \Rightarrow (x+4)(x+3)=0 \Rightarrow x=4$ or $ x=-3$, and these are the two solutions to our equation, so we are done.
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