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?

  • $\begingroup$ It might help to note that the (negative of the) solution you seek is the positive solution to $x=2^{-x/2}$. $\endgroup$ – Antonio Vargas Nov 24 '15 at 18:06
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    $\begingroup$ $-\frac{23}{30}$ is a very good approximation. $\endgroup$ – Peter Nov 24 '15 at 18:07
  • $\begingroup$ Well $x^2$ is always positive so if there is a negative solution it would be x = -y where y is positive and $y^2 = 1/2^y$. So y < 1. If g(x) = $y^2 - 1/2^y$ then g(x) is continuous. g(0) = -1. g(1) = 1/2. So there is a solution in there. $\endgroup$ – fleablood Nov 24 '15 at 18:11
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    $\begingroup$ @fleablood by a negative solution, I meant the negative value of x for which $x^2=2^x$ $\endgroup$ – user217339 Nov 24 '15 at 18:16
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    $\begingroup$ If you look for a "standard notation" for the negative solution, you might want to check out the Lambert W function. Using it, the negative solution is something like $x=-2W((\ln 2)/2)/\ln 2$. $\endgroup$ – mickep Nov 24 '15 at 18:45

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

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


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

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.

  • $\begingroup$ Doesn't answer how it is approximated in the first place. $\endgroup$ – Nihilist_Frost Nov 24 '15 at 18:18
  • $\begingroup$ See my comment above for an approximation. $\endgroup$ – Peter Nov 24 '15 at 18:18

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


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