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.

$$
e^y+e^{-y}=2
$$
Letting $u = e^y$, this becomes
$$
u + \frac 1u = 2
$$
Multiplying both sides by $u$:
$$
u^2 + 1 = 2u
$$
That's just a quadratic equation.
