Value of y slog(base (e^(pi/2))( y) = y

[Ivars]

I was strugling to get in grips with infinite pentation of other base than e but failed.

So my question to experts:

Which y would satisfy the equation in the subject of the thread?

My guess is e^(-pi), so that:

e^(pi/2) [5] ( - infinity) = e^(-pi).

Ivars

[GFR]

Thank you, Ivars. Let us see.

@ Andydude.

Could you please check the coordinates of the common intersection, for x < 0, and for b = e^(Pi/2), with your powerful slog and sexp machines, of:
- y = b # x = b-tetra-x;
- y = [base b]slog x, the inverse of the previous one;
- y = x, principal diagonal ?

The intersections of the three "tails" for x < 0 should correspond to b-penta(-oo).
But, I might be wrong.

GFR

[Ivars]

Is this very difficult or not interesting?
I still have not acquired software to be able to do it myself one day. I will proceed analytically, but that might take years

Ivars

[bo198214]

Is this very difficult or not interesting?

The value must be somewhere around -2 (far from your guess of \( e^{-\pi} \)) considering this picture showing the intersection of \( \text{slog}_{e^{\pi/2}} x \) with \( x \).
   
I guess it is not symboblically expressable with \( e \), \( i \) and \( \pi \).

But it is not exactly -2, because \( \text{slog}_b (x) > -2 \) for all real \( x \).

[Ivars]

Thanks!

Seems rather symmetrical, this value.

Ivars