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Homework Help: Solving w e^x

  1. May 26, 2005 #1
    how do i solve
    [tex]e^x+e^{3x}-1=0[/tex]
    for x?
     
  2. jcsd
  3. May 26, 2005 #2
    to make it simple let y=e^x

    thus giving y^3 + y - 1 = 0
    y(y^2 + 1) = 1

    i looked at this quickly and i don't think that you would get a solution for x.
     
    Last edited: May 26, 2005
  4. May 26, 2005 #3

    HallsofIvy

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    Let y= ex and the equation becomes y3+ y- 1= 0. Can you solve that?
     
  5. May 26, 2005 #4
    how can the answer be x=0?
    [tex]e^{x}+e^{3x}-1=0[/tex]
    [tex]e^{0}+e^{3\times{0}}-1=0[/tex]
    [tex]1+1-1=0[/tex]
    [tex]1=0[/tex]??
     
  6. May 26, 2005 #5
    the answer cannot be zero, i edited my post (sorry about that)
     
  7. May 26, 2005 #6
    no prob, but is there really no solution?
     
  8. May 26, 2005 #7
    well it makes sense to me because this is a cubic function right? how did you go about solving for y?
     
  9. May 26, 2005 #8
    i didn't, i don't know how to
     
  10. May 26, 2005 #9

    Zurtex

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    There is 1 real answer and 2 initial complex answers, being non-simple roots of the cubic equation they all look quite nasty.
     
  11. May 26, 2005 #10
    could someone explain how to find the real answer?
     
  12. May 26, 2005 #11

    dextercioby

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    Check out Mathworld's page on the cubic equation.

    Daniel.
     
  13. May 26, 2005 #12
    ok, thanks
     
  14. May 27, 2005 #13
    you can set
    [tex] y=e^x[/tex]
    then the equation will become [tex] y+y^3-1=0 [/tex]
    now you can creat the term [tex] y^2 [/tex],then minus the same term.
     
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