I would like to know how can I solve the following exponential equation for $x$: $$\exp\left(\frac{n_1}{x}\right) + \exp(n_2) + \exp\left(n_3 - \frac{n_4}{x}\right) - \exp(n_5) = 0$$ where $n_1$, $n_2$, $n_3$, $n_4$, and $n_5$ are constants.

Many thanks in advance!

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    $\begingroup$ I suspect you'll have to solve numerically, if there is even a solution. Do you know the $n_i$'s? $\endgroup$ – Adrian Keister Aug 8 at 14:37
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    $\begingroup$ For some integer values of the $n_i$, the equation can be multiplied by $e^{nx}$ for some $n$ and then solved as a polynomial in $e^x$, though in general this will not yield a closed-form solution. $\endgroup$ – Connor Harris Aug 8 at 14:39
  • $\begingroup$ Yes, I know $n_i$'s. So what is the best way to solve it? There is no possibility for a closed-form solution? $\endgroup$ – Igor Dakic Aug 8 at 14:41
  • $\begingroup$ The killer is $n_3$ Without that, you can get an equation of the form $a^x=b$ and take logs. $\endgroup$ – Ross Millikan Aug 8 at 14:47
  • $\begingroup$ the value for $n_1/n_4$ depends on the values I use, but it is probably greater than 1 and less than 4. $\endgroup$ – Igor Dakic Aug 8 at 14:47

Let $y=\exp(1/x)$ and $n_i'=\exp(n_i)$ and multiply both sides by $y^{n_4}$ to get


which is a trinomial in $y$, which cannot be algebraically solved in general. It can, of course, be solved numerically.

  • $\begingroup$ Nice answer! How can you prove that your equation cannot be algebraically solved? Thanks. $\endgroup$ – manooooh Aug 8 at 15:25
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    $\begingroup$ Abel-Ruffini theorem e.g. $y^5-y+1=0$ cannot be solved. $\endgroup$ – Simply Beautiful Art Aug 8 at 15:28
  • $\begingroup$ Thanks a lot! How can I solve it numerically? $\endgroup$ – Igor Dakic Aug 12 at 12:53
  • $\begingroup$ @IgorDakic There are plenty of root-finding algorithms you can see for yourself. $\endgroup$ – Simply Beautiful Art Aug 12 at 12:58
  • $\begingroup$ Thanks! Just one last question - it is possible to do an approximation of this equation and have some formulation for $x$? It does not have to be exact. $\endgroup$ – Igor Dakic Aug 13 at 8:51

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