Rich, I thought I had a promising idea re finding a unique flow for e^x (and if it works, it works equally for a^x, I think, at least for a > e^(1/e), and maybe all a <> 1 in (0,oo). So I was eager to hear your reponse. To repeat (tersely this time): Assume a > (e^(1/e) (though perhaps unnecessarily) and consider f_a(x) := a^x, x real. Conjugating with 1/x, use g_a(x) := a^(-1/x) for x > 0. To find *its* flow, first find the flow of g_(a,c)(x) := a^(-1/x) + c, which has a fixed point x_c > 0, satisfying 0 < g_(a,c)'(x_c) < 1. Now use our usual formula to get a unique flow for g_(a,c) (using only x > x_c). Letting c decrease to 0, I hope the limit of these flows is an analytic flow for a^(-1/x). Finally, conjugate the flow with 1/x again to recover the flow for a^x. (Which, if it works, would work for x < 0, but this can be easily extended to all x by applying log_a, etc.) What thinkst thou? --Dan