Solving tetration for base 0 < b < e^-e

[mike3]

Why not use the entire? Because the behavior is not consistent. It seems stranger to imagine a function that has singularities for only real bases \( b \geq e^{-e} \), then suddenly it becomes free of them and entire for all \( b < e^{-e} \). It's two wildly different analytical behaviors and that doesn't make much sense. Ideally it would be nice to be able to interpret the tetration at \( 0 < b < e^{-e} \) to be what you'd get if you did an analytical continuation in the base from \( b > e^{-e} \) through the complex plane. Finally, at the integer towers of this base, we are still dealing with real numbers: when we take the log of \( ^0 b = 1 \) for \( 0 < b < e^{-e} \) to get \( ^{-1} b \) as 0, we are still using a real logarithm of a real number to a real base. So why not continue using this principal real logarithm for the rest of the tet function at this base?