If tanh x = operatorname{sech} y = frac{3}{5} | vedclass

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(D) Given $	anh x = \frac{3}{5}$ $\Rightarrow \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{3}{5}$ $\Rightarrow \frac{e^{2x} - 1}{e^{2x} + 1} = \frac{3}{

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$6$

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(D) Given $	anh x = \frac{3}{5}$ $\Rightarrow \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{3}{5}$ $\Rightarrow \frac{e^{2x} - 1}{e^{2x} + 1} = \frac{3}{5}$.$5e^{2x} - 5 = 3e^{2x} + 3$ $\Rightarrow 2e^{2x} = 8$ $\Rightarrow e^{2x} = 4$ $\Rightarrow e^x = 2$.Given $\operatorname{sech} y = \frac{3}{5}$ $\Rightarrow \frac{2}{e^y + e^{-y}} = \frac{3}{5}$ $\Rightarrow \frac{2e^y}{e^{2y} + 1} = \frac{3}{5}$.$10e^y = 3e^{2y} + 3 \Rightarrow 3(e^y)^2 - 10e^y + 3 = 0$.Let $t = e^y$,then $3t^2 - 10t + 3 = 0$ $\Rightarrow (3t - 1)(t - 3) = 0$ $\Rightarrow t = 3$ or $t = \frac{1}{3}$.Since $e^{x+y} = e^x \cdot e^y$,we have $e^{x+y} = 2 	imes 3 = 6$ or $e^{x+y} = 2 	imes \frac{1}{3} = \frac{2}{3}$.Since $e^{x+y}$ is an integer,$e^{x+y} = 6$.

If $\tanh x = \operatorname{sech} y = \frac{3}{5}$ and $e^{x+y}$ is an integer,then $e^{x+y} =$

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