=Mike Stay [...] the problem: given a function (+) of two variables x,y, find a function p() such that p(x) + p(y) = p(x (+) y) Somehow they got from x (+) y = xy to p'(x) = 1/x and then integrated to get p = log.

Mike, I would be quite interested in seeing this paper when you locate it. From Melzak[*] I learnt what I style "Weierstrass's Elliptic Prison": "A function f(x) is said to possess an algebraic addition formula if there is a polynomial P...such that P( f(x+y), f(x), f(y) ) = 0 ...The theorem of Weierstrass states that the only [such] functions are elliptic functions and their special degenerate cases such as trigonometric, hyperbolic and exponential functions." So exo-elliptics spawn exotic additions (+) (née plus ultra<;-) [*] Z. A. Melzak, Companion to Concrete Math. 1973 p56--but no citation to Weierstrass (not that I'd be able to follow it<;-)