Solving hyperbolic functions I have 2 questions with regards to solving of hyperbolic functions. I have presented my current solutions to the best of my ability.  
Q1: Show that the real solution $x$ of $\tanh(x) = \operatorname{csch}(x)$ can be written in the form $x=\ln(u)+ \sqrt{u},$ where $u$ is to be determined.  
My attempt: write $\tanh x =\operatorname{csch}(x)$ as 
\begin{align*}
\dfrac{\sinh(x)}{\cosh (x)}&= \dfrac{1}{\sinh(x)}\iff \\
\sinh^{2}(x)&=\cosh(x)\iff \\
\cosh ^{2}(x)-1&= \cosh(x) \iff\\
 \cosh ^{2}(x)- \cosh(x)-1&=0 \iff \\
\cosh(x)&= \dfrac{1\pm\sqrt{5}}{2}.
\end{align*}
Writing $\cosh(x) =\dfrac{e^{x}+e^{-x}}{2}$, we have $ \dfrac{1+\sqrt{5}}{2}=\dfrac{e^{x}+e^{-x}}{2} \iff e^{2x}-(1+\sqrt{5})e^{x}+1=0$. Am I on the right track? This is where I am stuck because by applying the quadratic formula to solve for $e^{x}$ yields a double square root.  
Q2: Solve $\cosh(4x)+4\cosh(2x)-125=0$.  
My attempt: By using the identities $\cosh(4x)= \cosh^{2}(2x)+ \sinh^{2}(2x)$ and $\sinh^{2}(2x)=1+\cosh^{2}(2x)$ and substituting into the original equation and simplifying, we obtain: $$\cosh^{2}(2x)+ 2\cosh(2x)-62=0.$$ 
Solving, we obtain $\cosh(2x)=-1+3\sqrt{7}$ or $\cosh(2x)=-1-3\sqrt{7}$. Like the above problem, I am stuck but am I on the right track?   
 A: You're on the right track: the quadratic formula tells you that
$$
e^x=\frac{1+\sqrt{5}\pm\sqrt{(1+\sqrt{5})^2-4}}{2}=
\frac{1+\sqrt{5}\pm\sqrt{2(1+\sqrt{5})}}{2}
$$
If you set $u=\frac{1+\sqrt{5}}{2}$, then you get either
$$
e^x=u+\sqrt{u}
$$
or
$$
e^x=u-\sqrt{u}
$$
On the other hand
$$
u-\sqrt{u}=\frac{u^2-u}{u+\sqrt{u}}=\frac{1}{u+\sqrt{u}}
$$
So the first solution is
$$
x=\ln(u+\sqrt{u})
$$
and the second solution is
$$
x=-\ln(u+\sqrt{u})
$$
The positive real solution of your equation is of the stated form. Notice that
$$
\frac{\sinh(-x)}{\cosh(-x)}=-\frac{\sinh x}{\cosh x}
\qquad
\frac{1}{\sinh(-x)}=-\frac{1}{\sinh x}
$$
so any positive solution is accompanied by a negative one.
In both problems 1 and 2, the negative solution for $\cosh x$ must be discarded.
A: If
$\cosh(x) = y$,
then,
as stated,
$e^x+e^{-x} = 2y$
or
$e^{2x}-2ye^x+1 = 0$.
Solving,
$e^x
=\dfrac{2y\pm\sqrt{4y^2-4}}{2}
=y\pm\sqrt{y^2-1}
$
or
$x
=\ln(y\pm\sqrt{y^2-1})
$.
If
$y^2-1 = y$
(i.e.,
$y=\dfrac{1\pm\sqrt{5}}{2}
$),
then
$x=\ln(y\pm\sqrt{y})$.
