If operatorname{sech}^{-1} x = log 2 and oper | vedclass

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(B) Given $\operatorname{sech}^{-1} x = \log 2$. By definition,$\operatorname{sech}^{-1} x = \cosh^{-1} (\frac{1}{x}) = \log 2$. Thus,$\cosh^{-1} (\fr

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$\frac{1}{20}$

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(B) Given $\operatorname{sech}^{-1} x = \log 2$. By definition,$\operatorname{sech}^{-1} x = \cosh^{-1} (\frac{1}{x}) = \log 2$. Thus,$\cosh^{-1} (\frac{1}{x}) = \log 2$,which implies $\frac{1}{x} = \cosh(\log 2)$. Since $\cosh(	heta) = \frac{e^{	heta} + e^{-	heta}}{2}$,we have $\frac{1}{x} = \frac{e^{\log 2} + e^{-\log 2}}{2} = \frac{2 + \frac{1}{2}}{2} = \frac{5/2}{2} = \frac{5}{4}$. So,$x = \frac{4}{5}$. Given $\operatorname{cosech}^{-1} y = -\log 3$. By definition,$\operatorname{cosech}^{-1} y = \sinh^{-1} (\frac{1}{y}) = -\log 3$. Thus,$\frac{1}{y} = \sinh(-\log 3) = -\sinh(\log 3)$. Since $\sinh(	heta) = \frac{e^{	heta} - e^{-	heta}}{2}$,we have $\frac{1}{y} = -(\frac{e^{\log 3} - e^{-\log 3}}{2}) = -(\frac{3 - 1/3}{2}) = -(\frac{8/3}{2}) = -\frac{4}{3}$. So,$y = -\frac{3}{4}$. Therefore,$x + y = \frac{4}{5} - \frac{3}{4} = \frac{16 - 15}{20} = \frac{1}{20}$.

If $\operatorname{sech}^{-1} x = \log 2$ and $\operatorname{cosech}^{-1} y = -\log 3$,then $(x + y) = $

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