Algebraic Solution for Arcsech(x) = ?

[ChaoticLlama]

I need to determine an algebraic form for arcsech(x) = ?

So far what I've come up with is as follows:

$\L\<br /> \begin{array}{l}<br /> y = \frac{2}{{e^x + e^{ - x} }} \\ <br /> <br /> y = \frac{2}{{e^x + e^{ - x} }}\left( {\frac{{e^x }}{{e^x }}} \right) \\ <br /> <br /> y = \frac{{2e^x }}{{e^{2x} + 1}} \\ <br /> <br /> ye^{2x} - 2e^x + y = 0 \\ <br /> \end{array}$

Need to continue solving for x, thanks for any suggestions.

 

[HallsofIvy]

Let z= e^2x and you equation becomes
$yz^2- 2z+ y= 0$
Use the quadratic formula to solve for z and then 2x= ln z.

 

[arildno]

Set $u=e^{x}$
Thus, you get the equation in u:

$$yu^{2}-2u+y=0$$

1.Can you solve that for u, remembering that u>0?
2. Solve afterwards for x!