andydude Wrote:I just wanted to clarify that my super-logarithm solution only works for \( x > e^{1/e} \) even though my pdf says \( x > 1 \). I think that the reason for this is that tetration for bases between 1 and \( e^{1/e} \) has an upper limit (due to the convergence of \( {}^{\infty}x \)), and thus the super-logarithm has a limited domain over which it is defined.Are you sure, what goes wrong for bases \( b<\eta \)?
I mean it is clear that the domain of the superlog is bounded above by the lower fixed point of \( b^x=x \), which I will call \( \beta(b) \) and which is \( \beta(b)=W(-\log(b))/(-\log(b)) \).
So \( \text{slog}_b(x) \) is allowed only for \( x<\beta(b) \) because \( \lim_{n\to\infty}{}^nb=\beta(b) \).
What goes wrong for those arguments with your superlog?