I have been thinking about this equation:
$$x^2=2^x$$
I know there is two integer solutions: $x=2$ and $x=4$. But there also is a negative solution, that is approximately $x=-0.77$.
$$(-0.77)^2=0.5929$$ $$2^{(-0.77)}=0.5864...$$
Can we find this negative solution exactly?
On
The solutions for the equation $x^{a}=a^{x}$ are given by $x=-\frac{a}{\log a}W\left(-\frac{\log a}{a}\right)$, where $W(\cdot)$ is the Lambert W-function, i.e., the inverse function of $f(W)=We^{W}$. Then we get $x_{1}=2,x_{2}=4, x_{3}\approx-0.7\bar{6}$.
P.S: To obtain the value $x_{3}$ we need to consider $x<0$, we get $x=-\frac{a}{\log a}W\left( \frac{\log a}{a}\right)$ in our case $a=2$ give $x_{3}=-\frac{2}{\log 2}W\left( \frac{\log 2}{2}\right)\approx -0.7\bar{6}$.
$x$ is a solution of the equation $x^2 = 2^x$ if : $$\dfrac{\ln|x|}{x} = \dfrac{\ln 2}{2}$$ from the graph of the function : $$f(x) = \dfrac{\ln|x|}{x} - \dfrac{\ln 2}{2}$$
we deduce that there are only $3$ solutions $x = 2$, $x = 4$ and $x = -0.767$.