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I have this problem that I need to solve:

$$\frac{350}{1-e^{-9k}} = \frac{116}{1-e^{-870k}}$$

I have been trying to solve for $k$ for a week now and I can't seem to get the right answer. This is probably very simple for many of you, but I keep on getting the natural log of a negative number and I don't know how to deal with it.

Thanks for any help or hints!

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  • $\begingroup$ Is $k$ real? Integer? Complex? $\endgroup$ – Bobson Dugnutt Sep 30 '16 at 20:07
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    $\begingroup$ Hint: let $x = e^{-3k}$. Then you get a polynomial equation in $x$ to solve for $x \gt 0$ (but it's not pretty). $\endgroup$ – dxiv Sep 30 '16 at 20:08
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HINT:

$$\frac{350}{1-e^{-9\text{k}}}=\frac{116}{1-e^{-870\text{k}}}\Longleftrightarrow\frac{350}{116}=\frac{1-e^{-9\text{k}}}{1-e^{-870\text{k}}}$$

Now, let $x=e^{-3\text{k}}$ and $\frac{350}{116}=\frac{175}{58}$:

$$\frac{175}{58}=\frac{1-x^3}{1-x^{290}}$$


With WolframAlpha, I found a numerical solutions for $x$:

$$x\approx-0.996295082210765802862270776699665801755821153020031$$

So, now when $x<0$, $\text{k}$ can not be real.

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  • $\begingroup$ Thank you so much! I'll work on it now! $\endgroup$ – riley lyman Sep 30 '16 at 20:17
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    $\begingroup$ @nileylyman Note that this will give you a solution $x \approx -0.9963$, but that does not translate to a real solution for $k$ since $x$ is negative. $\endgroup$ – Alexis Olson Sep 30 '16 at 20:19
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I think the Newton's method is suitable for same as equation.

rearrange the equation to get

$$y=234-350e^{-870k}+116e^{-9k}$$ and use the following to find the roots $$k_{n+1}=k_n-\frac{y(k)_n}{y(k)'_n}$$

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  • $\begingroup$ The problem is that there is no real root. $\endgroup$ – Claude Leibovici Oct 1 '16 at 3:18

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