How Do You Solve (1+x)e^-x = 0.5 for x?

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Homework Help Overview

The discussion revolves around solving the equation (1+x)e^-x = 0.5 for the variable x. Participants are exploring methods to rearrange and analyze the equation, which involves both exponential and algebraic components.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants suggest taking the natural logarithm of both sides and applying logarithmic properties. There is a discussion about the feasibility of solving the equation analytically, with some expressing doubt about the possibility of an analytical solution.

Discussion Status

The conversation is ongoing, with various approaches being discussed. Some participants have provided insights into potential methods, such as the Lambert W function, while others are questioning the correctness of the problem setup and the steps taken so far.

Contextual Notes

There is uncertainty regarding the analytical solvability of the equation, and some participants note the need for careful manipulation to apply certain mathematical functions. Additionally, there are indications of misinterpretations of the original problem by some contributors.

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

[tex]\ln(a/b)=\ln a - \ln b[/tex]

And remember that [tex]\ln e = 1 = \mbox{cancels each other out}[/tex]
 
Last edited:
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:
I can't seem to go anywhere from your last step, is the problem written down correctly?
 
do you have to solve this analytically?
 
It's certainly impossible to solve analytically.
 
nicksauce said:
It's certainly impossible to solve analytically.

yes, that's why I asked the OP ;)
 
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
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
llemes4011 said:
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.
 
  • #12
sorry, i also didn't read the original question right *sigh*
 
  • #13
llemes4011 said:
sorry, i also didn't read the original question right *sigh*
It's ok, no worries.
 

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