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So, I have the function $$f(x)=\frac{2^x-2^{-x}}{2}.$$ I tried finding the inverse function the usual way I do, but I guess I'm stuck with these degrees. So far, I've come to this form $$y=\frac{2^{2x}-2^0}{2^{x+1}}.$$ When I used Wolfram Alpha Inverse Function Finder, it gave me this solution $$f^{-1}(x)=\log\left(x-\sqrt{x^2+1}\right)/\log(2).$$ I have no idea how it god those logs, please help me :D

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  • $\begingroup$ Maybe try to get a different expression for $f$ by taking the second derivative and solving the resulting second order ODE. $\endgroup$ – walcher Sep 3 '13 at 8:58
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Hint: $$\frac{2^x - 2^{-x}}{2} = y \qquad \Longleftrightarrow \qquad (2^x)^2 - 2y \cdot 2^x - 1 = 0.$$

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Note that $$f(x)=\frac{e^{x \log2}-e^{-x \log2}}{2}=\cosh (x \log2)=y, $$ so that $$x=\frac{1}{\log 2} \text{arccosh }y. $$

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