Using Natural Logarithms to solve for x

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To solve for x in terms of y using natural logarithms, the equation y = (e^(10x) + e^(-10x)) / (e^(10x) - e^(-10x)) is manipulated by multiplying both sides by the denominator. After simplifying, the equation is rearranged to isolate terms involving e^(20x). By factoring and applying the natural logarithm, the solution is derived as x = ln((y + 1) / (y - 1)) / 20. This approach effectively utilizes logarithmic properties to solve the equation. The final expression provides the relationship between x and y.
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Homework Statement


Use Natural Logarithms to solve for x in terms of y

y = \frac{e^{10x}+e^{-10x}}{e^{10x}-e^{-10x}}


Homework Equations


I am not too sure.


The Attempt at a Solution


I multiplied both sides by the denominator first.
Then I multiply by an LCD of e^{10x}
I end up with y(e^{20x})-y = (e^{20x})+1

EDIT: I don't really know where to go from here. Any direction would be greatly appreciated.
 
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From were you ended up, you could proceed by putting all terms of the form e^{20x} to one side of the equation. Then factor it out and take the logarithm.
 
micromass said:
From were you ended up, you could proceed by putting all terms of the form e^{20x} to one side of the equation. Then factor it out and take the logarithm.

Absolutely. Thank you so much for the help. I ended up with

x=\frac{ln\frac{y+1}{y-1}}{20}
 
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Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

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