I have been trying to solve the following relation for t:

$$exp(r_1*t)-exp(r_2*t) = K*(exp(r_1*t_a)-exp(r_2*t_a))$$

In this case, the following assumptions hold:

  • t is positive, real, and non-zero. This is the quantity to find.
  • t_a is a constant, real, positive, and non-zero.
  • r_1 and r_2 are non-zero and constant.
  • K is real, positive and non-zero, additionally K cannot be greater than 1. (When K is 1, we have a special case where t = t_a.)

The closest answers I could find to this problem are here:

Solve equation $\exp(ax)+\exp(bx)=1$

How do i solve $e^{ax}-e^{bx}=c$ for $x$?

Is there any possibility of an analytical solution in this case? Any hints on how it could be proven that there is no analytical solution? My big doubt is that I can plot the parent function (this is the simplified expression for one of the parameters in the function) and find the solution by analyzing the parent function. In practicality, t actually has 2 solutions, but I take the larger one. My intuition tells me there could be an analytical solution, but I can't find a way to fully prove whether or not it exists.


Everything on the right side is constant, so you're solving $$ e^{r_1 t} + e^{r_2 t} = c $$ where $c$ is some constant. Of course you need $c > 0$ for there to be a real solution. If $x = e^{r_1 t}$ and $p = r_2/r_1$, this is

$$ x + x^p = c$$

If $p$ is rational, this is equivalent to solving a polynomial (but except in a few cases, the solutions can't be expressed in radicals). For general $p$, there will be no closed-form solutions. However, there may be series solutions. Of course, for $c=2$, $x=1$ is a solution. If $c$ is near $2$, there is a series for a solution in powers of $c-2$:

$$ x =1+ \frac{c-2}{p+1} -{\frac {p \left( p-1 \right) }{2\, \left( p+1 \right) ^{3}}} \left( c-2 \right) ^{2}+{ \frac {p \left( p-1 \right) \left( {p}^{2}-p+1 \right) }{3\, \left( p +1 \right) ^{5}}} \left( c-2 \right) ^{3}-{\frac {p \left( p-1 \right) \left( 6\,{p}^{4}-13\,{p}^{3}+22\,{p}^{2}-13\,p+6 \right) }{ 24\, \left( p+1 \right) ^{7}}} \left( c-2 \right) ^{4}+{\frac {p \left( p-1 \right) \left( 4\,{p}^{6}-14\,{p}^{5}+33\,{p}^{4}-38\,{p} ^{3}+33\,{p}^{2}-14\,p+4 \right) }{20\, \left( p+1 \right) ^{9}}} \left( c-2 \right) ^{5}+ \ldots$$

  • $\begingroup$ Hi Rob! Thanks for the well-thought out answer. Do you have any references that I could do more reading on? $\endgroup$ – Aquiles Parodi Feb 1 '17 at 22:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.