What is the tenth superroot of $e$?

Assuming that the infinte tetraetion

$$ x^{x^{x^{....}}}=e $$

exists (which is indeed the case in the sense of a limit, see here for more detail), the limiting value is given by

$$ x_*^e=e $$

or

$$ x_*=e^{1/e}\approx1.44466786 $$

which should serve as an extremly good approximation for the solution of $$x\uparrow\uparrow10=e$$

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Update:

The real value seems to be

$$x\approx 1.46396$$

so the relative error using the infinte approximation

is

$$ \frac{|x-x_*|}{x}\approx0.0131 $$

which is pretty awesome regarding the simplicity of this approximate solution

Considering that even $x^x=a$ doesn't come with a formula for $x$ either, I think it is not unreasonable to search for the result by traditionnal means like dichotomy, especially since the tetration is an increasing function in the considered interval.

Also since $x\uparrow\uparrow n=e$ imposes strong bounds on $x$ in $[1,2]$ as you stated else it would diverge quickly, we are somehow in the ideal range for the $pow$ function accuracy.

For large $n$, the infinite approximation given by tired would work fine, and for small $n$, it would not be a big deal for a computer to calculate $x\uparrow\uparrow n$ with required accuracy.