How to solve $e^{ax}+e^{bx}+e^{cx}+d=0$ 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.
$$e^{ln(y)a}+e^{ln(y)b}+e^{ln(y)c}+d=y^a+y^b+y^c+d=0$$

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$\large\tt Hint:$
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.
