TITLE: Negative solution to $x^2=2^x$
QUESTION [7 upvotes]: Just out of curiosity I was trying to solve the equation $x^2=2^x$, initially I thought there would be just the two solutions $x=2$ and $x=4$, but wolfram shows that the two equations intersect at not 2 but 3 locations, the third being a negative value of $x$. The third solution isn't obvious like the other two, so I just have a few questions about the negative solution. Is it rational? is it commonly represented with a greek letter? If it is irrational is there a way to approximate it?

REPLY [3 votes]: So we have $x^2 = 2^x$. Taking the square root of both sides and assume the solution is negative gives $x=-\sqrt{2}^x$. We can then establish a recursive sequence, $x_n = -\sqrt2^{x_{n-1}}$. Assuming that this converges gives us the answer, $x=-\sqrt2 ^{-\sqrt2 ^ {-\sqrt2 ^\cdots}}$. After five iterations, we get $$x\approx-0.76961847524.$$ Substituting the answer back in we get, $$2^{-0.76961847524} = 0.58657257487 \approx 0.59231259743 =(-0.76961847524)^2.$$