An (im)possible closed form solution to $x^{x^{-3}}=2^{x^{-3}-2^{x-3}}$

Question

Solve in real numbers $$x^{x^{-3}}=2^{x^{-3}-2^{x-3}}$$

My attempt

To be honest this is not an actual attempt, but that is all I could do. That question was shared in a math group. $x^{-3}$ together with $x-3$ is mind blowing. The first thing I did to answer the question was to set up graphics .

There is exactly one real root and this root appears to be positive. As a result of numerical approximation $x\approx1.28237148900997...$

I could not approach the question algebraically. The question is beyond my knowledge. But out of curiosity I tried a lot of rational and irrational numbers, for example $x=\frac{\sqrt {41}}{5}$. I could not find a number matching the number $x\approx 1.2823$

Does the equation have a solution known as a closed form, or can it ?

Answers from all perspectives are appreciated.

Answer 1

Solution

Let $x=2w$: $$x^{x^{-3}}=2^{x^{-3}-2^{x-3}}\iff-\log_2\left(\frac x2\right)=2^{x-3}x^3\iff -\log_2(w)=4^ww^3$$

and notice the graphs cross at $(w,w)$

so we set one equal to $w$ and use Lambert W:

$$w=-\log_2(w)\iff w2^w=1\\\boxed{x=2\frac{\operatorname W(\ln(2))}{\ln(2)}=2e^{-\operatorname W(\ln(2))}}$$

Equation Derivation

The solution is not fully rigorous where it says the graphs cross at $(w,w)$, so here is probably how the equation’s creator made it, for a clearer perspective. Start with a transcendental equation and solve for $w$:

$$w2^w=1\implies \color{blue}w=-\log_2(w)$$

Also, square both sides and multiply by $w$ in the starting equation:

$$w2^w=1\iff (w 2^w)^2w=1^2w\iff4^ww^3=\color{blue}w$$

Since the two equations marked in blue both equal $w$, equate them and rearrange:

$$-\log_2(w)=4^ww^3\iff w^{w^{-3}}=2^{-4^w}$$

Finally, substitute $w=\frac x2$:

$$w^{w^{-3}}=2^{-4^w} \iff x^{x^{-3}}=2^{x^{-3}-2^{x-3}}$$

Answer 2

Considering that you are looking for the zero of function $$f(x)=\log\Big(x^{x^{-3}}\Big)-\log\Big(2^{x^{-3}-2^{x-3}}\Big),$$ you could define the solution to be the infinite limit of the first iterate of a Newton-like method of order $n$ starting with $$x_0=2\log(2)$$ The formulae will be nasty, very long to type but totally explicit.

I give you the very first (for Newton method) $$x_{(2)}=2 L-\frac{2 L \left(2^{2 L} L^4+\log (L)\right)}{2^{2 L+1} L^6-3 \log (L)+1}\qquad \text{where} \qquad L=\log(2)$$ Computing the next ones and converting them to decimals

$$\left( \begin{array}{ccc} n & x_{(n)} & \text{method} \\ 2 & \color{red}{1.2}6372070923 & \text{Newton}\\ 3 & \color{red}{1.282}31922203 & \text{Halley}\\ 4 & \color{red}{1.282}45476924 & \text{Householder}\\ 5 & \color{red}{1.28237}340841 & \text{no name}\\ 6 & \color{red}{1.282371}25177 & \text{no name}\\ 7 & \color{red}{1.2823714}7462 & \text{no name}\\ \cdots & \cdots & \cdots\\ \infty & \color{red}{1.28237148901}& \text{solution}\\ \end{array} \right)$$