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.
 A: 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.
A: Following the method of ONG SEE HAI HCI  , we have $A = 2x+1$
then:
$$x^{(x-1)^2}=x^{x^2+2-(2x+1)}=2x+1$$
then $x^{x^2+2-A} = A $ with only solution $x=\sqrt{A}$.
This is because $x^{x^2+2-A}$ is increasing.
Then $x=\sqrt{A} \rightarrow x^2 = 2x+1 \rightarrow x = 2+ \frac{1}{x} \rightarrow x-\frac{1}{x} = 2$
A: A hint to the first problem.
Prove that for $x\geq2$ the function $f(x)=(x-1)^2\ln{x}-\ln(2x+1)$ increases.
Indeed, for $x\geq2$ we obtain:
$$f'(x)=2(x-1)\ln{x}+\frac{(x-1)^2}{x}-\frac{2}{2x+1}>0.$$
Thus, since $f(2)<0$ and $\lim\limits_{x\rightarrow+\infty}f(x)=+\infty,$ 
we see that our equation has an unique root on $[2,+\infty).$
Now, easy to check that $1+\sqrt2$ is a root.
Indeed, $$(1+\sqrt2)^{(1+\sqrt2-1)^2}-2(1+\sqrt2)-1=(1+\sqrt2)^2-3-2\sqrt2=0.$$
The second equation we can solve by the similar way. I got that $4$ is a root.
A: I think I understand what @Michael Rozenberg is trying to say now. Basically, he views the equation $f(x)=g(x)$ as the point(s) of intersection between the two functions $y=g(x)$ and $y=f(x)$. For example, for the second problem, to solve $x^{x-x^2+13}=x^2-12$, he says that we are essentially finding the points of intersection of $2$ functions, $f(x)=x^{x-x^2+13}$ and $g(x)=x^2-12$. But under such an interpretation, we have to take the domains of the functions into consideration, which means that there might be solutions to the original solution that were being omitted. 
So here's how to proceed: First note that $D_f=x \in \mathbb{R^+_{0}}$, and $D_g=\mathbb{R}.$ Now, $x^2-12=-(x-x^2+13)+x+1$, so let $x-x^2+13=A$. Thus, the original equation reduces to: $x^A=-A+x+1$, $x \geq 0$. Then, treating $A$ as a constant, we have the solution to this new equation to be $A=1$; this can be seen by inspection. It is easy to prove that no other values of $A$ will give us equality:
Let $h(x)=x^A+A-x-1,x\geq0$. When $x=A$, we have $h(x)=0$. Also, $h'(x)=Ax^{A-1}-1$. If $A>1$, $h'(x)=Ax^{A-1}-1 > x^{A-1}>x^0-1>0,$ where the second last inequality follows since $x$ is non-negative. Thus, $h(x)$ is strictly increasing and $\geq 0$ $\forall \ x \in \mathbb{R^+_{0}} \Rightarrow \ f(x)>g(x) \ \forall \ x \in \mathbb{R^+_{0}}$. Likewise, if $A<1, h'(x)=Ax^{A-1}-1<x^{A-1}-1<x^0-1<0$ implies $h(x)$ is strictly decreasing and $ \leq 0$$\forall \ x \in \mathbb{R^+_{0}} \Rightarrow \ f(x)<g(x) \ \forall \ x \in \mathbb{R^+_{0}}$.
Thus, given $A=1$, let $x-x^2+13=1 \Rightarrow x=-3 $ or $x=4$, but we reject the former solution since we have to take into account the domain of $f(x)$, $D_f$.
