06/10/2008, 09:31 PM
There is one possibility, but I'm not sure. The tetrational function \( {}^{x}(e^{1/e}) \) as \( x\to\infty \) converges to e, so there is no way to solve \( {}^{x}(e^{1/e})=10 \) for real x, but I'm not sure if this makes new numbers, or if its just the plain old complex numbers.
Andrew Robbins
Andrew Robbins

