# How to solve difficult exponential equation

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.

• I suspect you'll have to solve numerically, if there is even a solution. Do you know the $n_i$'s? – Adrian Keister Aug 8 at 14:37
• 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. – Connor Harris Aug 8 at 14:39
• Yes, I know $n_i$'s. So what is the best way to solve it? There is no possibility for a closed-form solution? – Igor Dakic Aug 8 at 14:41
• The killer is $n_3$ Without that, you can get an equation of the form $a^x=b$ and take logs. – Ross Millikan Aug 8 at 14:47
• the value for $n_1/n_4$ depends on the values I use, but it is probably greater than 1 and less than 4. – 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
$$y^{n_1+n_4}+(n_2'-n_5')y^{n_4}+n_3'=0$$
which is a trinomial in $$y$$, which cannot be algebraically solved in general. It can, of course, be solved numerically.
• Abel-Ruffini theorem e.g. $y^5-y+1=0$ cannot be solved. – Simply Beautiful Art Aug 8 at 15:28
• 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. – Igor Dakic Aug 13 at 8:51