Finitist Wrote:I'm guessing that the reason it's harder to find a "unique" solution, if one exists at all, is that, unlike addition and multiplication, exponentiation is neither commutative nor associative; hence there isn't the same level of "regularity" in the operations preceding tetration as there is in those preceding exponentiation.
I see it similarly. The usual uniqueness condition for the multiplication \( f(x)=bx \) is \( f(x+y)=f(x)+f(y) \) and for the exponentiation \( f(x)=b^x \) is \( f(x+y)=f(x)f(y) \). However we can not continue on that path for tetration/superexponential \( f \) as surely \( f(x+y)\neq f(x)^{f(y)} \) for most integer \( x,y \) (except if we switch to arborescent numbers, which I investigated in my phd thesis).