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There are $2$ negative solutions for $2^x=x^{-2}$ but also a positive one. The positive one is approximately $0.766664$ ($6$ s.f. because of desmos) but I can't find a nondecimal expression for this solution anywhere.

Cusp Connoisseur
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Dominik Trory
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    For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User Oct 01 '24 at 22:19
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    It's not much, but I can tell you that $x\approx 0.766664695962123$. (Using Desmos) – Cusp Connoisseur Oct 01 '24 at 22:22
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    This question is similar to: Solve for $x$ in $a^x=x^b$. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. The function in that question has these representations/properties – Тyma Gaidash Oct 01 '24 at 22:23
  • By using the logarithm, you get $$ x \ln 2 = -2 \ln x $$ Solve for $\ln x$ on the right side: $$ \ln x = \frac{- \ln 2}{2} x $$ Exponentiate: $$ x = (e^{-\ln 2/2})^x = (2^{-1/2})^x $$ Then this old post of mine applies with $$ b = 1 \qquad c = 0 \qquad a = e^{-\ln 2/2} = \frac{1}{\sqrt 2} $$ – PrincessEev Oct 01 '24 at 22:52
  • We see, your equation is a polynomial equation of more than one algebraically independent monomials ($2^x,x$) and with no univariate factor. We therefore don't know how to solve the equation for $x$ by rearranging by applying only finite numbers of elementary functions (elementary operations) we can read from the equation. It is therefore likely that non-elementary Special functions are required. – IV_ Oct 03 '24 at 08:31

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As far as I (and Mathematica) are aware, there is no closed form solution, at least if you are looking for something in terms of the familiar functions and operations like roots, trigonometric functions, and the like.

To get a symbolic solution, you need to invoke what's called the Lambert W function. This function, which is actually multi-valued (the same input value can and will lead to multiple outputs) is defined by the relation:

$$w e^{w} = z \qquad \Longrightarrow \qquad w = W(z)$$

More information can be found on Wikipedia and its associated references.

As for a precise closed-form solution, you are looking for $\frac{2 W \big( \frac{\log(2)}{2}\big)}{\log(2)}$.

P.S. I don't know how far along in your mathematics career you are, but in case you are fairly new to this world, let me give some advice. Although many of the problems given to you and books and by your teachers will have fairly nice solutions in terms of familiar functions, practically speaking, these special functions will prove more common. You should get used to these situations where functions are not defined in terms of some nice geometry, but instead by satisfying some very special relation.

In fact, the $\log$ function is really the first taste of that fact.

DiracComb16796
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  • It'll be nice if you show the way how you get the precise answer? It's pretty easy. Otherwise a good answer – Gwen Oct 01 '24 at 23:07