I need to determine an algebraic form for arcsech(x) = ?(adsbygoogle = window.adsbygoogle || []).push({});

So far what I've come up with is as follows:

[itex]\L\

\begin{array}{l}

y = \frac{2}{{e^x + e^{ - x} }} \\

y = \frac{2}{{e^x + e^{ - x} }}\left( {\frac{{e^x }}{{e^x }}} \right) \\

y = \frac{{2e^x }}{{e^{2x} + 1}} \\

ye^{2x} - 2e^x + y = 0 \\

\end{array}

[/itex]

Need to continue solving for x, thanks for any suggestions.

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# Homework Help: Hyperbolic Trig Inverses

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