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

• Jun 8th 2010, 10:46 AM
ice_syncer
ln(x)=e^-x find x
ln(x)=e^(-x)

find x... :(
• Jun 8th 2010, 12:39 PM
Quote:

Originally Posted by ice_syncer
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
• Jun 8th 2010, 07:35 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 ....")
• Jun 9th 2010, 03:11 AM
ice_syncer
how did you think of the initial shot ie x = 1.6
• Jun 9th 2010, 04:56 AM
Quote:

Originally Posted by ice_syncer
how did you think of the initial shot ie x = 1.6

$\displaystyle e^{-x}$ 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.
• Jun 9th 2010, 10:37 AM
ice_syncer