How to solve an equation like $e^{ax}+e^{bx}+e^{cx}+d=0$ (i.e. to write $x=...$) where $a,b,c,d$ are fixed non-zero real numbers.
I have tried assuming that $x=ln(y)$ for $y>0$ but it goes nowhere.
$$e^{ln(y)a}+e^{ln(y)b}+e^{ln(y)c}+d=y^a+y^b+y^c+d=0$$
On
Sketch your function and locate a point close to the solution. Then apply Newton method.
For example, let me set a=1.5, b=2.6, c=3.7 and d=-10^7. The solution is between x=4 and x=5. Let us start Newton iterations at x=4; then the successive iterates are 4.7298, 4.5267, 4.33902, 4.3575, 4.3540, 4.3540.
You can easily find bounds playing with inequalities. In the case I chose, the solution is such that Exp[3.7 x] < 10^7 which means x < 4.3562 which will be a very good starting point.
On
Thinking more about this problem, I suppose that we could build a kind of recurrence relation in the spirit of what we would do if only one exponential was present. Let me suppose that a < b < c and that these constants are "sufficiently" different from each other. Then, the problem is basically governed by Exp[c x]. So, let us write
xnew = Log[- d - Exp[a xold] - Exp[b xold] ] / c
and start using xold = 0. This gives xnew = 4.35624 then 4.35397 then 4.35398.
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\large\tt Hint:$
We can set $a \leq b \leq c.\quad$ $d$ must be negative ($d \equiv -\expo{\mu}$). The equation becomes $$ {\expo{ax} \over \verts{d}} + \verts{d}^{b/a - 1}\pars{{\expo{ax} \over \verts{d}}}^{b/a} + \verts{d}^{c/a - 1}\pars{{\expo{ax} \over \verts{d}}}^{c/a} = 1\,,\qquad a \not= 0\,,\quad d < 0 $$
$$ z + z_{b}\,z^{\beta} + z_{c}\,z^{\gamma} = 1 \quad\mbox{where}\quad \left\vert% \begin{array}{rcl} z & \equiv & {\expo{ax} \over \verts{d}} \quad\yy\quad x = {\ln\pars{\verts{d}z} \over a} \\[1mm] z_{b} & \equiv &\verts{d}^{b/a - 1}\,,\quad\beta\equiv {b \over a} \\[1mm] z_{c} & \equiv& \verts{d}^{c/a - 1}\,,\quad\gamma\equiv {c \over a} \\[2mm] \quad a \not= 0\,,&& d < 0 \end{array}\right. $$
Since the equation solution requires $\quad 0 < z < 1,\quad$ $0 < z_{b}z^{\beta} < 1\quad$ and $\quad 0 < z_{c}z^{\gamma} < 1,\quad$ it's obvious that a numerical candidate is the bisection method.
When $a = 0$ (with $b \leq c$), we can repeat the above procedure. Whenever $a = b = 0$ and $c \geq 0$ the analysis is trivial.