Slog(Exponential Factorial(x))

[tommy1729]

what i wonder if how basechange(e,x) behaves as a function of slog(x).

basechange(e,x) = a0 + a1 slog(x) + a2 slog(x)^2 + ...

regards

tommy1729

Well we know that 

slog(x^x^x^…) - X

And 

slog(2^3^…^x) - X

Are approximed by

Slog( ln(x) x + ln(ln(x)) ) 

Or 

Slog( ln(x-1) x + ln(ln(x-2)) )

Which explains impliciet how the base Change formula should work and why

At least for c^oo tetration and real bases x > eta.

This suggesties that a Taylor of slog might not be the best idea.

Maybe slog(f(x)) + slog(f(x))/(x - eta) 

For a Taylor f or so.

Hmmm

Regards

Tommy1729