This problem is related to a time constant (herein called $x$) possessed by an article heated in a constant-temperature oven. Below I have simplified the presentation of the problem so as to concentrate on the math aspects. I do not want to get off onto discussing times and temperatures.

Is there a direct symbolic solution for $X$ in the following equation? i.e, $X=$ _________
(Even a converging series would be nice.)

$J\left(\exp\left(\frac ax\right) -1\right) = K\left(\exp\left(\frac bx\right) -1\right)$

Assume (for laboratory use):

$J, K, a, b$ are real;

$a > b > 0$;

$|K| > |J| $

In any given experiment, $K$ and $J$ have the same polarity (either both are positive, or both are negative)

One set of numerical values:

$J=18; K=52; a=3.8; b=2.25; x= 1.824551$

I converted the equation to an infinite series, but all of the $x$'s were in the denominator:

$$... + (Kb^n-Ja^n)/(x^nn!) + \ldots = 0 \;\;\;(n=1, 2, 3, \ldots)$$

I am not a mathematician. I came across this problem while working in a laboratory in 1983, and have worked on it off and on since then.


In the most general case, an equation such as $$f(x)=J \left(e^{a/x}-1\right)-K \left(e^{b/x}-1\right)$$ does not have any analytic solution (the exception would be if $a$ was a small multiple of $a$ $2,3,4$ because in this case, a simple change of variable transforms the equation to a polynomial).

If the solution $x$ was large, we could replace $x$ by $\frac{1}{y}$ and expand (just as you did) as a Taylor series built at $y=0$. This would lead to an infinite expansion like $$f(y)=y (a J-b K)+\frac{1}{2} y^2 \left(a^2 J-b^2 K\right)+\frac{1}{6} y^3 \left(a^3 J-b^3 K\right)+O\left(y^4\right)$$ Limited to second order, the approximate solution would then be $$y=\frac{1}{x}=-\frac{2 (a J-b K)}{a^2 J-b^2 K}$$ Using your numbers, we should then find that $x=-\frac{37}{1080}\simeq -0.0342593$ which is just awful compared to what you give. Expansion in series could be only useful for estimation purposes under a lot of conditions.

All of that being said, only numerical methods should be used to solve your equation. A simple root finding method is Newton which, starting from a "reasonable" guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac {f(x_n)} {f'(x_n)}$$ For the case you gave, let us use $x_0=2$. The iterates will then be $1.75961$, $1.81760$, $1.82447$, $1.82455$ which is the solution for six significant figures.


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