A question on the equation $^qx=2$

Question

Given the equation $$^qx=2$$ with $q\gt3$ where $^qx$ means the 'tetration' operation on $x$, my question is: is it possible to find a value for $q$ for which the solution $x$ of the equation is a rational number? Thanks in advance.

Answer 1

The upper bound for your number is: $b=1.476684337357869947089235585 \;\;\; b \uparrow \uparrow 3 = 2$

If you extend tetration to the real numbers, than any $b>\sqrt{2}$ and less than the answer I gave would have $b \uparrow \uparrow x = 2$ for some number x>=3 and less than infinity. For example, I used Newton's method for these results.

$$b=1.47668433735787 \;\;\; b \uparrow \uparrow 3 = 2\;\; \text{upper bounds}$$ $$b=1.4584946676581 \;\; b \uparrow \uparrow 3.5=2$$ $$b=1.4466014324299 \;\; b \uparrow \uparrow 4=2$$ $$b=1.4326948056602 \;\; b \uparrow \uparrow 5=2$$

In the limit, if $b=\sqrt{2}$, then the limit as n goes to infinity,

$$\lim_{n \to \infty} \sqrt{2} \uparrow \uparrow n = 2$$

$$\text{if} \;b<\sqrt{2}, \;\; \text{then} \;\; \lim_{n \to \infty} b \uparrow \uparrow n < 2$$