Finding inverse cosh I am trying to find $\cosh^{-1}1$ I end up with something that looks like $e^y+e^{-y}=2x$. I followed the formula correctly so I believe that is correct up to this point. I then plug in $1$ for $x$ and I get $e^y+e^{-y}=2$ which, according to my mathematical knowledge, is still correct. From here I have absolutely no idea what to do as anything I do gives me an incredibly complicated problem or the wrong answer.

It may be more helpful to consider the significant hyperbolic identities first. We have in general:    
$\small \begin{array} {rcllll} 
1)&  \exp(z) &=& \cosh(z) + \sinh(z) \\  
2)&   1 &=&  \cosh(z)^2 - \sinh(z)^2 \\ 
&&&  \implies \\
3)&\sinh(z) &=& \pm \sqrt{\cosh(z)^2-1} & \text{ using 2)}\\
4)& \exp(z)&=& \cosh(z) \pm \sqrt{\cosh(z)^2-1}  & \text{ using 1) and 3)}\\
 \end{array} $      
Now the given problem is to find another expression for $\small y=\cosh^{-1}(x)$ which means $\small x = \cosh(y) $
We use 4) and insert our current y for the general z to get    
$\small \begin{array} {rcllll} 
5)& \exp(y)&=& \cosh(y) \pm \sqrt{\cosh(y)^2-1}  & \text{ using 4)}\\
6)& \exp(y)&=& x \pm \sqrt{x^2-1}  & \text{ inserting x for } \cosh(y)\\
7)& y&=& \log(x \pm \sqrt{x^2-1} ) & \\
8)& \cosh^{-1}(x)&=& \log(x \pm \sqrt{x^2-1} ) &\text{ inserting } \cosh^{-1}(x) \text{ for } y \\
9)& \cosh^{-1}(1)&=& ??? \\
 \end{array} $      
Now 8) can be used as a new, general hyperbolic identity like that in the list from 1) to 4) and 9) is your remaining little to-do ...
