How do I solve for N in 50N + 50 <= 2^N?
Not sure if relevant anymore, but hey, it's solvable but not with elementary functions. First things first, we know that 2
N is convex. Since 50*0 + 50 > 2
0, we know that 50N + 50 intercepts 2
N exactly twice due to the latter being convex. These intercepts occur for one positive N and one negative N. Call them N
1 and N
2 respectively. Thus the region for which 50N + 50 <= 2^N holds will be N <= N
2 or N >= N
1. The get the actual values, you can either solve them numerically...
50N + 50 = 2N <-> 50N + 50 = eN ln 2 <-> (N + 1)e-N ln 2 = 1/50 <-> (-N ln 2 - ln 2)e-N ln 2 = - (ln 2)/50 <-> (-N ln 2 - ln 2)e-N ln 2 - ln 2 = - ((ln 2)/50) e- ln 2
<-> -N ln 2 - ln 2 = Wk(- ((ln 2)/50) e- ln 2) <-> N= -Wk(- ((ln 2)/50) e- ln 2)/(ln 2) - 1 <-> N1 = -W-1(- ((ln 2)/50)e- ln 2)/(ln 2) - 1 , N2 = -W0(- ((ln 2)/50) e- ln 2)/(ln 2) - 1
Where Wk(x) is the kth branch of the Lambert W function. Only W0(x) and W-1(x) are real-valued. W0(x) is defined for x in [-1/e,∞) and is positive for x > 0, where as W-1(x) is defined for x in [-1/e,0) and is always negative. In fact, W-1(x) <= W0(x) for the common domain.
Though it's highly unlikely to for one to solve it symbolically outside of specialist settings.