Solving for 'a' in a Complex Equation

[Denyven]

Homework Statement

Solve for a

7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a

Homework Equations

7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a

The Attempt at a Solution

7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a

multiply by 2 on both sides

14=a(e^\frac{4}{a} + e^\frac{-4}{a}) + a

this is where i am stuck

is it even possible?

 

[LCKurtz]

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.

 

[thebigstar25]

Homework Statement

Solve for a

7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a

Homework Equations

7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a

The Attempt at a Solution

7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a

multiply by 2 on both sides

14=a(e^\frac{4}{a} + e^\frac{-4}{a}) + a

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

 

[thebigstar25]

hint: don't multiply with 2, and use your knowledge of Hyperbolic functions.. things may get easier from there ..

 

[HallsofIvy]

hint: don't 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!

\frac{e^{\frac{4}{a}}+ e^{-\frac{4}{a}}}{2}= cosh(\frac{4}{a})
but there is still a problem with that "a" outside the cosh.

 

[Denyven]

Hmmm Alright. So I now have
7=a cosh(\frac{4}{a})+a
I guess I would move a over and divide by a to get
\frac{7-a}{a}=cosh(\frac{4}{a})
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!

 

[LCKurtz]

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

<br /> \frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a\hbox{ and } 7<br />