# 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.