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?
Best Answer
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.