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.
$$e^{ln(y)a}+e^{ln(y)b}+e^{ln(y)c}+d=y^a+y^b+y^c+d=0$$
 A: $\newcommand{\+}{^{\dagger}}%
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$\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.
A: 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.
A: 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.  
