Negative solution to $x^2=2^x$

Question

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?

Answer 1

Suppose , $gcd(a,b)=1$ , $x=\frac{a}{b}$ and $x^2=2^x$

We have $$a^2=b^2\times 2^{a/b}$$

implying

$$a^{2b}=b^{2b}\times 2^a$$

This is impossible, if $gcd(a,b)=1$ and $b>1$. The negative solution is obviously not an integer.

If $x$ is irrational algebraic, then $2^x$ is transcendental, but $x^2$ is not.

So, $x$ , the negative solution, must be a transcendental number.

Answer 2

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, $$2^{-0.76961847524} = 0.58657257487 \approx 0.59231259743 =(-0.76961847524)^2.$$