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.

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 $x=\sqrt{A} \rightarrow x^2 = 2x+1 \rightarrow x = 2+ \frac{1}{x} \rightarrow x-\frac{1}{x} = 2$
