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

[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.

 

[chaoseverlasting]

Take log on both sides and go from there.

 

[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 $$\ln e = 1 = \mbox{cancels each other out}$$

 

[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

 

[rocomath]

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

 

[malawi_glenn]

do you have to solve this analytically?

 

[nicksauce]

It's certainly impossible to solve analytically.

 

[malawi_glenn]

It's certainly impossible to solve analytically.

yes, that's why I asked the OP ;)

 

[HallsofIvy]

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)= xe^x- and may require manipulation of the equation to put it in that form.

 

[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))

 

[rocomath]

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))

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

 

[llemes4011]

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

 

[rocomath]

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

It's ok, no worries.