ln(x)=e^(-x)
find x...
Hi ice_syncer,
you could use Newton's Method for finding roots.
$\displaystyle f(x)=e^{-x}-lnx$
The solution to $\displaystyle lnx=e^{-x}$
is the root of $\displaystyle f(x)=0$
Take an initial shot at the solution as x=1.6
Then
$\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=1.6-\frac{e^{-1.6}-ln(1.6)}{-e^{-1.6}-\frac{1}{1.6}}=1.275767$
A few more iterations would close in on the root, approximately 1.31
$\displaystyle e^{-x}$ is a decreasing function, while $\displaystyle lnx$ is increasing.
$\displaystyle e^{-1}=0.368$
$\displaystyle ln1=0$
$\displaystyle e^{-2}=0.135$
$\displaystyle ln2=0.693$
Hence these cross between x=1 and x=2.
You could choose any value of x in the vicinity as a starting point, say x=1.5 or so. Choosing x=1 or x=2 would also suffice.
The attachment shows Newton's method homing in on the solution.