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 Value of y slog(base (e^(pi/2))( y) = y Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/03/2008, 04:40 PM (This post was last modified: 02/03/2008, 07:40 PM by 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 Member Posts: 174 Threads: 4 Joined: Aug 2007 02/03/2008, 11:54 PM 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 Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/22/2008, 01:21 PM 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 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/26/2008, 10:45 AM (This post was last modified: 02/26/2008, 10:54 AM by bo198214.) Ivars Wrote: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 Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/26/2008, 01:13 PM Thanks! Seems rather symmetrical, this value. Ivars « Next Oldest | Next Newest »

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