Fast method to solve transcendental equation for a range of parameters? I have an equation of the form
$t + e^{Ax} + e^{Bx} = 0$
This is a transcendental equation, and I would use a Newton-Raphson algorithm or uniroot in R, to solve for $x$.
I am interested in finding the solution to this equation for a range of values of $t$.
Do I need to repeat the same algorithm for every value of $t$, or is there any way in which I can use the solution for a particular $t$ to calculate the solution for other values of $t$?

Edit: In my specific problem I have $A > 0$ and $B<0$.
 A: Here is a solution that transforms the initial problem with 2 parameters into a 1 parameter issue, easier to tackle.
Set $C=B/A$ (a negative number because $A>0$ and $B<0$).
Let us make a (bijective !) change of variable
$$u=e^{Ax} \ \ \iff \ \ x=\frac{1}{A}\text{ln}(u)\tag{1}$$
Important remark : $u>0$.
The initial equation
$$x+e^{Ax}+e^{Bx}=0 \tag{2}$$
is thus transformed into
$$\underbrace{t+u+u^C}_{f(u)}=0\tag{3}$$
or, geometrically, into the intersection of
$$\begin{cases}y&=&u^C& \ \ \text{(curve with asymptotes)} \\y&=&-u-t& \ \ \text{line with slope} \ -1\end{cases}\tag{4}$$
Fig. 1 illustrates (4).
Three cases can occur (due to the fact that the straight line hits the $y$ axis in $-t$) :

*

*either zero solutions when $t<t_0$ for a certain $t_0$.


*(exceptionally) one solution when $t=t_0$,


*2 solutions when $t=t_0$,

and in this case which one do you desire ?
I have assumed here that we are looking for the root with the highest magnitude.

The choice of the initial value is directed by the diagram : if we take the initial value $u=-t$, for big values of $-t$, we will be already rather close to the final value.
The following Matlab program applying Newton iteration is very efficient :
A=1.;B=-2.5;C=B/A;
t=-3;
u=-t, % initialization
for k=1:4, % 4 iterations are usually enough
   u=u-(t+u+u^C)/(1+C*u^(C-1)), % Newton iteration step u = u-f(u)/f'(u)
end;
x=(1/A)*log(u), % solution x=x(t)
t+exp(A*x)+exp(B*x), % checking that this value is (almost !) zero


Fig. 1 : (case  $A=1, B=-2.5, C=B/A=-2.5$. Horizontal axis is for $u$ variable). Curve with equation $y=u^C$ and two straight lines with resp. equations $y=-u-t_0$ and $y=-u-t$ for two different values of $t$ ($t=-3$ and $t=t_0 \approx 1.818$); Newton recurrence relationship with initial value $u_0=3$ gives final value $u_5=2.9321...$ out of which $x=1.0757...$.
