Eh... I *think* you meant what would be expressed in J as: 0 = 8 + (2^x) - 2^2^x
I'd probably try maybe a few hundred rounds of newton's method first, and see where that leads. But there's an ambiguity where the original expression (depending on the frame of reference of the poster) could have been intended to be: 0 = 8 + (2^x) + _2^2^x [if that is solvable, x might have to be complex] Thanks, -- Raul On Sun, Jan 28, 2018 at 5:25 PM, Skip Cave <[email protected]> wrote: > What is the best iterative way to solve this equation: > > (-2^2^x) + (2^x) +8 =0 > > > Skip Cave > Cave Consulting LLC > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
