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I'm trying to solve an equation, but can't find any fitting formula or algorithm.

$$e^{ax} + e^{b(x - t)} = 2, t > 0$$

At the moment, I'm using an approximate solution, but my intuition tells me there may be an analytical one, however, it seems that I don't know how to solve this equation by myself. Can you help me with this? Many thanks in advance.

EDIT 1: I'm looking for a solution for $x$.

EDIT 2: I hope I'm right at least here:

  • When $a = 0$ and $b = 0$, $x$ may be anything.
  • When $a = 0$ and $b \neq 0$, $x = t$.
  • When $a \neq 0$ and $b = 0$, $x = 0$.
  • When $a > 0$ and $b > 0$, the function on the left is increasing, and there's one $x > 0$.
  • When $a < 0$ and $b < 0$, the function on the left is decreasing, and there's one $x < t$.
  • When $a \neq 0$ and $b \neq 0$, and they have opposite signs, there're three possibilities:
    • if $f(\frac{bt}{b - a}) > 2$, there's no $x$;
    • if $f(\frac{bt}{b - a}) = 2$, there's one $x = \frac{bt}{b - a}$;
    • if $f(\frac{bt}{b - a}) < 2$, there're $x_1 < \frac{bt}{b - a}$ and $x_2 > \frac{bt}{b - a}$.
Irimitlad
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    I would be surprised if it had a closed form solution for general $a,b,t$. But I don't have a quick reference from the top of my head and proving that something is cannot be expressed in "closed form" can be hard to prove. Cf. here for comparison. – Thomas Preu Aug 14 '22 at 17:22
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    You are trying to solve for what? I can easily solve for $a$, or for $b$, or for $t$. Only solving for $x$ will be a problem. – GEdgar Aug 14 '22 at 18:31
  • @GEdgar Right, exactly, it's for $x$, this's why it's such a problem. – Irimitlad Aug 14 '22 at 18:50
  • @ThomasPreu Now I also think that there may be no a general analytical solution. Maybe, I was tricked by this equation's seeming simplicity. – Irimitlad Aug 14 '22 at 18:58
  • What are the signs of $a$ and $b$ ? This is a key point to clarify – Claude Leibovici Aug 15 '22 at 07:45
  • @ClaudeLeibovici The point is their signs can be anything. – Irimitlad Aug 15 '22 at 08:40
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    It is perhaps worth noting that if a=nb, then setting $z=e^{bx}$ and $c=e^{-bt}$, the equation simplifies to $z^n+cz=2$. Even in this case, the problem seems intractable for large values of n. – Aaron Aug 15 '22 at 09:31

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