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Solve for a

  • Thread starter Denyven
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  • #1
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Homework Statement



Solve for a

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

Homework Equations



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

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:

Answers and Replies

  • #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.
 
  • #3
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Homework Statement



Solve for a

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

Homework Equations



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

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?
when you multiplied with 2, you didnt multiply the last term, which should give 2a
 
  • #4
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hint: dont multiply with 2, and use your knowledge of Hyperbolic functions.. things may get easier from there ..
 
  • #5
HallsofIvy
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hint: dont multiply with 2, and use your knowledge of Hyperbolic functions.. things may get easier from there ..
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.
 
  • #6
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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!
 
  • #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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