Common factor out from a sum of exponential functions

Question

From the below equation, which is a sum of two exponential functions I would like to compute the common factor $n$ $$ d = \exp\left(\frac{-x}{n}\right)+\exp\left(\frac{-y}{n}\right)$$ Unfortunately, values $x, y$ are both significantly bigger than $n$, so I cannot use the approximation $e^x \approx 1 - x$, because it works only for small values of $x$. Is there any other way to compute value $n$, for fixed $d, x, y$ even approximately?

Answer 1

This will be possible only through a numerical method, like Newton's

Answer 2

In the most general case, the solution cannot be explicit and, just as vonbrand answered, numerical methods would be required.

The problem can be made quite simple since the solution will be such that $$\min \left(-\frac{x}{\log \left(\frac{d}{2}\right)},-\frac{y}{\log \left(\frac{d}{2}\right)}\right)\leq n\leq \max \left(-\frac{x}{\log \left(\frac{d}{2}\right)},-\frac{y}{\log \left(\frac{d}{2}\right)}\right)$$ which is a good news.

We can make the problem easier setting $n=\frac 1t$ and then consider that the equation is $$d=e^{-xt}+e^{-yt}$$ but depending on the values of $d,x,y$, the corresponding function could be quite stiff. We can make it better considering instead $$\log(d)=\log\Big(e^{-xt}+e^{-yt}\Big)$$ So, you can use Newton method searching for the zero of $$f(t)=\log\Big(e^{-xt}+e^{-yt}\Big)-\log(d)$$ which will be close to a straight line.