Fix r = 0.4, and for x > 0, define
f(x) = log(x^(r^x)) = r^x log x,
g(x) = log(r^(x^r)) = x^r log r.
We want to solve f(x)=g(x).
If x >= 1, then f(x) >= 0 > g(x), so there are no solutions there.
Now, for 0 < x < 1, |r^x| and |log x| are both decreasing, but f(x) < 0, so f(x) is increasing, from -infinity to 0. On the other hand, x^r is increasing, but log r < 0, so g(x) is decreasing, from 0 to log(r).
Therefore there is a unique solution to f(x)=g(x), which must be x=r.
|
|
Posted by Eigenray
on 2008-03-18 18:51:49 |
solution
