1. ## ln(x)=e^-x find x

ln(x)=e^(-x)

find x...

2. Originally Posted by ice_syncer
ln(x)=e^(-x)

find x...
Hi ice_syncer,

you could use Newton's Method for finding roots.

$f(x)=e^{-x}-lnx$

The solution to $lnx=e^{-x}$

is the root of $f(x)=0$

Take an initial shot at the solution as x=1.6

Then

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=1.6-\frac{e^{-1.6}-ln(1.6)}{-e^{-1.6}-\frac{1}{1.6}}=1.275767$

A few more iterations would close in on the root, approximately 1.31

3. Originally Posted by ice_syncer
ln(x)=e^(-x)

find x...
Please post the whole question (in particular the background to it that probably includes something like "Find the approximate solution of ....")

4. how did you think of the initial shot ie x = 1.6

5. Originally Posted by ice_syncer
how did you think of the initial shot ie x = 1.6
$e^{-x}$ is a decreasing function, while $lnx$ is increasing.

$e^{-1}=0.368$

$ln1=0$

$e^{-2}=0.135$

$ln2=0.693$

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.