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Background information that may help you answer:

Alright, so I'm working on a formula that posits that there are a unique pair of coordinates $(x_1, y_1)$ and $(x_2, y_2) $ on the hyperbolic cosine function $(\cosh(x))$ that have a given secant slope and a given secant distance to arc length ratio. Setting up a system of two equations, our first equation dealing with slope comes from the difference quotient:

$$\frac{\cosh(x_2)-\cosh(x_1)}{x_2-x_1}=m_{sec}$$

Our second is the distance between the two coordinates divided by the arc length of the hyperbolic cosine between the two points. The numerator is a simple distance formula:

$$\sqrt{[\cosh(x_2)-\cosh(x_1)]^2+[x_2-x_1]^2}=l_{sec}$$

Finding the denominator:

$$y=\cosh(x)$$

$$l_{arc}=\int_{x_1}^{x_2}\sqrt{\left(\frac{dy}{dx}\right)^2+1}$$

$$l_{arc}=\int_{x_1}^{x_2}\sqrt{\sinh^2(x)+1}$$

$$l_{arc}=\int_{x_1}^{x_2}\cosh(x)$$

$$\therefore l_{arc}=\sinh(x_2)-\sinh(x_1)$$

Thus, our two equations are:

$$\frac{\cosh(x_2)-\cosh(x_1)}{x_2-x_1}=m_{sec}$$

$$\frac{\sqrt{[\cosh(x_2)-\cosh(x_1)]^2+[x_2-x_1]^2}}{\sinh(x_2)-\sinh(x_1)}=\frac{l_{sec}}{l_{arc}}$$

Where $m_{sec}$, $l_{sec}$, and $l_{arc}$ are known constants. $$\\$$

Actual Problem:

Now here's where my problem is: I need to isolate either $x_2$ or $x_1$.

The first equation is clearly the easiest. Let's rewrite it as an exponential.

$$e^{x_2}+e^{-x_2}+e^{x_1}+e^{-x_1}=2m_{sec}(x_2-x_1)$$

Despite multiple approaches using logarithms,I've made very little progress in isolating either of the variables. I simply don't have enough knowledge of logarithms to isolate either variable.

If you prefer to help me out by giving me hints, that's fine. I've just laboured over this problem for a month now to no avail and suppose it's time I look for help.

EDIT: David provides a nice approach (upvote given), but I'm still interested in other approaches, specifically one working logarithms as I've outlined. I want to make substitution as painless as possible.

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HINT...

Writing $M=m_{sec}$ and $L=\frac{l_{sec}}{l_{arc}}$:

You could start by substituting the $\cosh $ terms from the first equation into the second so you get $$\sqrt{(M^2+1)(x_2-x_1)^2}=L(\sinh x_2-\sinh x_1)$$

Now substituting $(x_2-x_1)$ back in the first equation you have $$\pm\frac{\sqrt{M^2+1}}{M}(\cosh x_2-\cosh x_1)=L(\sinh x_2-\sinh x_1)$$ $$\Rightarrow\pm\frac{\sqrt{M^2+1}}{M}(2\sinh\frac{x_2+x_1}{2}\sinh\frac{x_2-x_1}{2})=L(2\cosh\frac{x_2+x_1}{2}\sinh\frac{x_2-x_1}{2})$$ $$\Rightarrow\tanh\frac{x_1+x_2}{2}=\pm\frac{LM}{\sqrt{M^2+1}}$$

From here you can get one of the x's in terms of the other and substitute back to the $l_{arc}$ equation...

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    $\begingroup$ Following the first arrow is the application of the hyperbolic equivalents of the addition formulae for trig functions, applying Osborne's Rule, of course. $\endgroup$ – David Quinn Dec 26 '15 at 19:44
  • $\begingroup$ I noticed this after I posted, my apologies for such a trivial question. I thoroughly appreciate your response, but I am going to wait to see if any method providing a simpler substitution arises before I select your answer and kill the question $\endgroup$ – Lanier Freeman Dec 26 '15 at 19:50

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