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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?

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  • $\begingroup$ I don't see the form $\ln(u)+\sqrt u$ in your proposal. But in fact, it is very unlikely, there must be a typo. $\endgroup$ – user65203 Jun 11 '18 at 12:57
  • $\begingroup$ Part 1 is trivial. Every real number can be written as $ f(u) = \ln u + \sqrt{u}$ with $ u > 0,$ because $f(u)$ is strictly increasing, $f(u) \rightarrow -\infty$ for $u \rightarrow 0$ and $f(u) \rightarrow \infty$ for $u \rightarrow \infty.$ $\endgroup$ – gammatester Jun 11 '18 at 13:00
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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.

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  • $\begingroup$ thanks for your help. For the second question, can you Kindly check if my workings are correct as I still can't seem to arrive at an answer. Likewise, I let $\cosh 2x = \dfrac{e^{2x}+e^{-2x}}{2}=-1+3\sqrt{7}$. Simplifying and using the quadratic equation I obtain an ugly expression. $\endgroup$ – Cleytus Jun 11 '18 at 13:50
  • $\begingroup$ You get $e^{4x}-2(3\sqrt{7}-1)e^{2x}+1=0$, which is solved by the quadratic formula in the same way: finally you get $2x=\ln(3\sqrt{7}-1+\sqrt{63-6\sqrt{7}})$ and its negative. $\endgroup$ – egreg Jun 11 '18 at 13:58
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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 $x=\ln(y\pm\sqrt{y})$.

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