How can I express h in terms of D for this problem?

[whatsoever]

Homework Statement

[PLAIN]http://img88.imageshack.us/img88/1679/problemhup.png

Homework Equations

The Attempt at a Solution

[PLAIN]http://img535.imageshack.us/img535/8075/solutionz.png
That is how far i have got, using that sinh(x)=1/2(e^(x)-e^(-x))
I have to express h in terms of D

 

[Mentallic]

Firstly let

$$\sqrt{\frac{2.09\cdot 10^6}{D}}=x$$

just to make things clearer and easier to write out. Now we have

$$0.5=\frac{e^{xh-10x}-e^{10x-xh}}{e^{xh}-e^{-xh}}$$

So we need to find h in terms of x (which is in terms of D, we can substitute back at the end)
Multiplying through by the denominator of the fraction, and then by [itex]2e^{xh}[/tex] remember that $e^x\cdot e^{-x}=1$ and [tex]e^{x}\cdot e^{x}=e^{2x}[/itex]
So we now have

$$e^{2xh}-1=2e^{2xh-10x}-2e^{10x}$$

and from here just rearrange, factorize out the exponents with h present, and solve from there using logs and such.

 

[whatsoever]

Seems like i have made a mistake sinh(x+y)=sinh(x).cosh(y)+sinh(y).cosh(x)
considering that and your help I've got this
[PLAIN]http://img571.imageshack.us/img571/6678/probp.png
but i have no idea what to do next

 

[Mentallic]

Seems like i have made a mistake sinh(x+y)=sinh(x).cosh(y)+sinh(y).cosh(x)

I haven't studied sinh myself, so I took your word for it. I like that it's similar to the sin(a+b) expansion

considering that and your help I've got this
[PLAIN]http://img571.imageshack.us/img571/6678/probp.png
but i have no idea what to do next

Again multiply through by e^xh and you can easily simplify things, such as the $$e^{2xh-10x}e^{-2xh-10x}$$ term

 

[whatsoever]

I haven't studied sinh myself, so I took your word for it. I like that it's similar to the sin(a+b) expansion
Again multiply through by e^xh and you can easily simplify things, such as the $$e^{2xh-10x}e^{-2xh-10x}$$ term

i've made a mistake when writing it its not $$e^{2xh-10x}e^{-2xh-10x}$$, its $$e^{2xh-10x}+e^{-2xh-10x}$$

 

[Mentallic]

Then multiply through by e^2xh. You'll get a quadratic in e^2xh, and if you can't see it, let u=e^2xh and treat other terms such as e^-10x as constants, then solve the quadratic in u, then substitute back.