# Homework Help: Solve for a

1. Apr 24, 2010

### Denyven

1. The problem statement, all variables and given/known data

Solve for a

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

2. Relevant equations

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

3. 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?

Last edited: Apr 24, 2010
2. Apr 24, 2010

### 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.

3. Apr 25, 2010

### thebigstar25

when you multiplied with 2, you didnt multiply the last term, which should give 2a

4. Apr 25, 2010

### thebigstar25

hint: dont multiply with 2, and use your knowledge of Hyperbolic functions.. things may get easier from there ..

5. Apr 25, 2010

### HallsofIvy

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.

6. Apr 25, 2010

### 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!

7. Apr 25, 2010

### LCKurtz

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

$$\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a\hbox{ and } 7$$

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook