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anonymous
 one year ago
help me find the natural logarithum of csch^1 ((1)/(2 sqrt(30)))
anonymous
 one year ago
help me find the natural logarithum of csch^1 ((1)/(2 sqrt(30)))

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arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Hyperbolic cosecant of x = csch x = 2/(e^x  e^x)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\(\large ln [csch^{1} (\frac{1}{2 \sqrt(30)})]\) ?? i think there may be a misunderstanding in the question or on my part which this might clarify \(\large cosech(y) = (\frac{1}{2 \sqrt(30)}) = \frac{2}{e^y  e^{y} }\) let \( (\frac{1}{2 \sqrt(30)}) = \alpha \) \(\large \alpha (e^y  e^{y}) = 2\) \(\large \alpha e^{2y}  2e^y  \alpha = 0\) let \(\large z =e^y \) \(\large \alpha z^2  2 z  \alpha = 0\) \(\large z = \frac{2 \pm \sqrt{4 + 4 \alpha^2}}{2 \alpha} = \frac{1 \pm \sqrt{1 + \alpha^2}}{ \alpha}\) \(\large y = ln [\frac{1 \pm \sqrt{1 + \alpha^2}}{ \alpha}] = ln [\frac{1}{\alpha } + \sqrt{1 + \frac{1}{\alpha^2}}]\) the last tweak should ensure we get the positive number for the log and so the question is this: isn't the log already in the inverse cosech or are you taking a further log?
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