Originally Posted by

**rain** My fault, I should have made this clearer,

Given that you can get down to this equation,

$\displaystyle e^{x} -e^{-x} = 2y$

Then multiplying all the terms (both on the left hand side and the right hand side) by $\displaystyle e^{x}$ gives

$\displaystyle e^{x}(e^{x} -e^{-x}) = e^{x}2y$

Then, by expanding the bracket out, you get

$\displaystyle e^{x}e^{x} -e^{-x}e^{x} = e^{x}2y$

Now, by standard rules of indices, $\displaystyle a^ba^c=a^{b+c}$ and $\displaystyle a^0 = 1$ we get $\displaystyle e^{-x}e^{x} = e^{-x+x} = e^{0} = 1$

Then $\displaystyle e^{x}e^{x} -1 = e^{x}2y$ which can be written as

$\displaystyle (e^{x})^2 -e^{x}2y -1 = 0$ [1] by collecting the $\displaystyle e^{x}$ terms and bringing the 2y term to the left hand side

Now, say you let W be $\displaystyle e^{x}$, just to show this is like a quadratic, you can rewrite [1] as $\displaystyle W^2 -W2y -1 = 0$ which is in the general formula for solving a quadratic equation, with roots at $\displaystyle W=\frac{(2y)\pm\sqrt{(2y)^2-4(-1)}}{2}

=\frac{2y\pm\sqrt{4y^2+4}}{2}

=\frac{2y\pm\sqrt{4(y^2+1)}}{2}$

$\displaystyle =\frac{2y\pm2\sqrt{y^2+1}}{2}$ (since $\displaystyle \sqrt{4} = 2$)

$\displaystyle =y\pm\sqrt{y^2+1}$ (by dividing by 2)

thus $\displaystyle W=e^{x}=y\pm\sqrt{y^2+1}$

and taking the log of both sides gives

$\displaystyle x = log(y\pm\sqrt{y^2+1})$ and the rest carries on above

*The formula for solving a quadratic equation in general is as follows in case you can't remember,

for $\displaystyle ax^2 +bx +c = 0$

then $\displaystyle x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$