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

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• June 8th 2010, 11:46 AM
ice_syncer
ln(x)=e^-x find x
ln(x)=e^(-x)

find x... :(
• June 8th 2010, 01:39 PM
Archie Meade
Quote:

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
• June 8th 2010, 08:35 PM
mr fantastic
Quote:

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 ....")
• June 9th 2010, 04:11 AM
ice_syncer
how did you think of the initial shot ie x = 1.6
• June 9th 2010, 05:56 AM
Archie Meade
Quote:

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.
• June 9th 2010, 11:37 AM
ice_syncer
.......ok... I'll read more about it, thanks!