2009-11-16, 11:02 PM
Horusmaster Wrote:let a=2^nIt's 2^(n^2), not (2^n)^2.
thus the equation a^2-a=5
solve a (do it urself i'm too lazy)
then n= log a/ log 2
Unless it is (2^n)^2, in which case that does work.

