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Homework Help: Rearranging (1+x)e^-x = 0.5

  1. Jan 6, 2008 #1
    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. jcsd
  3. Jan 6, 2008 #2
    Take log on both sides and go from there.
     
  4. Jan 6, 2008 #3
    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: Jan 6, 2008
  5. Jan 6, 2008 #4
    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
  6. Jan 6, 2008 #5
    I can't seem to go anywhere from your last step, is the problem written down correctly?
     
  7. Jan 6, 2008 #6

    malawi_glenn

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    do you have to solve this analytically?
     
  8. Jan 6, 2008 #7

    nicksauce

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    It's certainly impossible to solve analytically.
     
  9. Jan 6, 2008 #8

    malawi_glenn

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    yes, thats why I asked the OP ;)
     
  10. Jan 6, 2008 #9

    HallsofIvy

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    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.
     
  11. Jan 16, 2008 #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))
     
  12. Jan 16, 2008 #11
    That's wrong, read all the posts before yours.
     
  13. Jan 16, 2008 #12
    sorry, i also didn't read the original question right *sigh*
     
  14. Jan 16, 2008 #13
    It's ok, no worries.
     
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