What is the inverse function of $\alpha\mathrm{e}^{\beta x}+\gamma\mathrm{e}^{\delta x}$?

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I need to solve this equation for $x$: $\alpha\mathrm{e}^{\beta x}+\gamma\mathrm{e}^{\delta x}=\epsilon$

($\alpha$, $\beta$, $\gamma$, $\delta$, $\epsilon$ are real constants)

I'm only interested in real solutions.

Is such an equation solvable? Even Wolfram|Alpha refuses to solve trivialized versions of this equation: $\mathrm{e}^{\alpha x}+\mathrm{e}^{\beta x}=a$ or $\alpha^x+\beta^x=a$

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It isn't possible to find a general solution. Letting $y = e^x$, we have $$\alpha y^\beta + \gamma y^\delta = \epsilon$$ Even for some integer values of $\alpha, ...,\epsilon$, this could give us polynomials whose solutions cannot be expressed in terms of ordinary operations (addition, subtraction, multiplication, division, and roots). For example, the roots of $$y^5-4y=1$$ can't be expressed in terms of ordinary mathematical operations, since it's irreducible and it has three real roots and two complex roots, and thus has Galois group $S_5$. And so if we can't find a solution for this special case, certainly there won't be a general solution.

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Letting $z = \alpha e^{\beta x}/\epsilon$, and assuming the parameters are positive, the equation becomes $$ z + \dfrac{\gamma \epsilon^{\delta/\beta-1}}{\alpha^{\delta/\beta}} z^{\delta/\beta} - 1 = 0$$ which I'll write as $$ z + c z^p - 1 = 0 $$

This has a series solution in powers of $c$, that should converge for small $|c|$:

$$\eqalign{z &= \sum_{n=0}^\infty \dfrac{((-c)^n}{n!} \prod_{j=0}^{n-2} (np - j)\cr &= 1 - c + p c^2 - \dfrac{3p(3p-1)}{6} c^3 + \dfrac{(4p)(4p-1)(4p-2)}{24} c^4 + \ldots}$$

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As already said in comments and answers, in the most general case, equations such that $$\alpha\mathrm{e}^{\beta x}+\gamma\mathrm{e}^{\delta x}=\epsilon$$ will not show analytical solutions and numerical methods would be required. The exceptions would correspond to situations where the ratio $\frac \beta \delta$ or $\frac \delta \beta $ would be a small $(<4)$ whole number.

Assuming all constants to be positive, the function $$f(x)=\alpha\mathrm{e}^{\beta x}+\gamma\mathrm{e}^{\delta x}-\epsilon$$ varies very fast and locating the root could be difficult. On the other hand, considering $$g(x)=\log(\alpha\mathrm{e}^{\beta x}+\gamma\mathrm{e}^{\delta x})-\log(\epsilon)$$ could be much more convenient since the plot of the new function will look (more or less) to a straight line.

For illustration purposes, let us consider the case of $$f(x)=123 e^{\pi x}+456 e^{e x}-123456789$$ and $$g(x)=\log(123 e^{\pi x}+456 e^{ex})-\log(123456789)$$ In both cases, a plot of the function reveals that the solution is close to $x_0=4$. So, let us use Newton method starting at this value.

Using $f(x)$, the iterates will be $$\left( \begin{array}{cc} 1 & 4.3640459464006870 \\ 2 & 4.2648712725384999 \\ 3 & 4.2468522386110215 \\ 4 & 4.2463481388110046 \\ 5 & 4.2463477579278545 \\ 6 & 4.2463477579276373 \end{array} \right)$$ Using $g(x)$, the iterates will be $$\left( \begin{array}{cc} 1 & 4.2467859955218309 \\ 2 & 4.2463477592885205 \\ 3 & 4.2463477579276374 \end{array} \right)$$

Using $g(x)$ with $x_0=0$ gives as a first estimate $x_1=4.367$.