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Solve for a

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve for a

    [tex]7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a[/tex]

    2. Relevant equations

    [tex]7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a[/tex]

    3. The attempt at a solution

    [tex]7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a[/tex]

    multiply by 2 on both sides

    [tex]14=a(e^\frac{4}{a} + e^\frac{-4}{a}) + a[/tex]

    this is where i am stuck

    is it even possible?
     
    Last edited: Apr 24, 2010
  2. jcsd
  3. Apr 24, 2010 #2

    LCKurtz

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    There certainly won't be a nice closed formula for a and there may be no real value of a that solves it at all.
     
  4. Apr 25, 2010 #3
    when you multiplied with 2, you didnt multiply the last term, which should give 2a
     
  5. Apr 25, 2010 #4
    hint: dont multiply with 2, and use your knowledge of Hyperbolic functions.. things may get easier from there ..
     
  6. Apr 25, 2010 #5

    HallsofIvy

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    That's a good suggestion but I doubt things will get much easier!

    [tex]\frac{e^{\frac{4}{a}}+ e^{-\frac{4}{a}}}{2}= cosh(\frac{4}{a})[/tex]
    but there is still a problem with that "a" outside the cosh.
     
  7. Apr 25, 2010 #6
    Hmmm Alright. So I now have
    [tex]7=a cosh(\frac{4}{a})+a[/tex]
    I guess I would move a over and divide by a to get
    [tex]\frac{7-a}{a}=cosh(\frac{4}{a})[/tex]
    Now I don't know what to do. Is there some property of cosh in which I can pull something out or split something? We just started hyperbolic functions and I still know very little.
    Thanks very much for all the help so far!
     
  8. Apr 25, 2010 #7

    LCKurtz

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    There is no real valued solution. Here's a graph of the relevant portion of

    [tex]
    \frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a\hbox{ and } 7
    [/tex]



    graph.jpg
     
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