# Help solving: $exp(r_1*t)-exp(r_2*t) = K*(exp(r_1*t_a)-exp(r_2*t_a))$))

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