Once again, sorry!

Thinking of the negative <-1> iteration of the exponential, which (if I am not mistaken) gives the log, I just realized that we might have:

y = b[3]x = exp_b (x), which gives the following right-inverse:

x = b[3]\ y = log_b (y), which is the <-1> iteration of exp_b on y:

x = b[3]<-1> y = log_b (y).

Therefore, we might have the following relation between these operations:

b[3]<-1> x = b[3]\<1> x = b[3]\ x = log_b (x)

And (perhaps) in general, showing only the operators, we indeed have:

b[s]<-n> = b[s]\<n> = s-hexp_b <-n> = s-hlog_b <n> , with back and forward iterations.

Right? ... or ... wrong !!!

GFR

Thinking of the negative <-1> iteration of the exponential, which (if I am not mistaken) gives the log, I just realized that we might have:

y = b[3]x = exp_b (x), which gives the following right-inverse:

x = b[3]\ y = log_b (y), which is the <-1> iteration of exp_b on y:

x = b[3]<-1> y = log_b (y).

Therefore, we might have the following relation between these operations:

b[3]<-1> x = b[3]\<1> x = b[3]\ x = log_b (x)

And (perhaps) in general, showing only the operators, we indeed have:

b[s]<-n> = b[s]\<n> = s-hexp_b <-n> = s-hlog_b <n> , with back and forward iterations.

Right? ... or ... wrong !!!

GFR