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Is it possible to write $1-e^{-xy} = r(x)r(y)$ for some function $r$ where $x,y$ are positive real numbers. I was just wondering to try to express that quantity like that. I tried solving the equation by Brut force but was not able to make any impact. Any suggestion will be helpful.

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    $\begingroup$ Hint: consider $r(x)r(1)$ and $r(x)r(2)$; their quotient should be constant.... (Indeed, this addresses the more general problem $1-e^{-xy} = r(x)s(y)$.) $\endgroup$ – Greg Martin Jun 20 '18 at 6:04
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That would imply $1-e^{-x^2}=r(x)^2$ and then $r(x)=\sqrt{1-e^{-x^2}}$. Therefore $1-e^{-xy}=\sqrt{(1-e^{-x^2})(1-e^{-y^2})}$. But this is false: try $x=1$ and $y=2$.

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Do partial differenciation on the both sides with $x$, and then assume that let $x=y$, then integrate on both sides with $x$ with limits $[o,x]$ as we know $r[o]$ so we can get the $r[x]$.

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    $\begingroup$ Welcome to MSE! Please use MathJax for formatting. $\endgroup$ – Bill O'Haran Jun 20 '18 at 6:58

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