User blog comment:D57799/Some discoveries and guesses about uparrow notation/@comment-5307762-20150128141343

1. Yes, but no in exponentiation.

3. Based on an approximation of the function y=a upup x. I know it's ill-defined for non-integer a.

4.Same reason as 3.

6.

a up^3 b = a up^2 (a up^3 (b-1))

slog_a (a up^3 b) = a up^3 (b-1)

a up^3 (-2)

= slog_a (a up^3 (-1))          passes(-1,0)

= slog_a(0)          (the solution to:   a upup x=0)

= -1