Rearranging (1+x)e^-x = 0.5

1. Jan 6, 2008

Xeract

Sorry if this is a very simple question, I am trying to rearrange (1+x)e^-x = 0.5 for x, and just can't seem to get my head around it. Any tips would be greatly appreciated.

2. Jan 6, 2008

chaoseverlasting

Take log on both sides and go from there.

3. Jan 6, 2008

rocomath

Take the natural log of both sides and then use the properties of log to split the left side.

$$\ln(a/b)=\ln a - \ln b$$

And remember that $$\ln e = 1 = \mbox{cancels each other out}$$

Last edited: Jan 6, 2008
4. Jan 6, 2008

Xeract

This is where I've got to, thanks for the responses:

ln((1+x)/e^x) = ln0.5
ln(1+x)-x =ln0.5

Last edited: Jan 6, 2008
5. Jan 6, 2008

rocomath

I can't seem to go anywhere from your last step, is the problem written down correctly?

6. Jan 6, 2008

malawi_glenn

do you have to solve this analytically?

7. Jan 6, 2008

nicksauce

It's certainly impossible to solve analytically.

8. Jan 6, 2008

malawi_glenn

yes, thats why I asked the OP ;)

9. Jan 6, 2008

HallsofIvy

Staff Emeritus
In general, a problem involving "x" both in an exponent and not can only be solved using the "Lambert W function" which is defined as the inverse function to f(x)= xex- and may require manipulation of the equation to put it in that form.

10. Jan 16, 2008

llemes4011

I don't know if i'm doing this right, but here are my thoughts. if you factor out the first side you get xe^x+e^x=0.5 Then if you divide both sides by x you get e^x+e^x=0.5/x add your like terms on the left and you get 2e^x=0.5/x (i think that's right)
See if you can get it from there
((i just realized that this was from over a week ago AFTER i finished lol))

11. Jan 16, 2008

rocomath

That's wrong, read all the posts before yours.

12. Jan 16, 2008

llemes4011

sorry, i also didn't read the original question right *sigh*

13. Jan 16, 2008

rocomath

It's ok, no worries.