The number of real roots of an exponential equation with 4 parameters The equation of interest is:
$$x=k_0e^{ax}-k_1e^{bx}$$
where $k_0,k_1>0$ and $b>a$ are the parameters. It is possible to determine the number of the real roots of the equation when all the parameters are given specific values. However, I wonder, in general, how we can partition regions in the parameter space such that for each point in the same partition the corresponding equation has the same number of real roots.
I am sorry if the scope of the question is too broad and general and I expect that the answer to such question maybe really long, so I would be appreciated if anyone could just give me a hint or a reference of tools that I might need to grasp the general idea of how those parameters affect the number of real solutions.
 A: If $a=b$, then write $k_1 e^{ax}-k_2e^{bx}$ as $ke^{ax}$.
If $k$ and $a$ are opposite sign, or if either is 0, then $x=ke^{ax}$ has exactly one solution.
Otherwise, $$x=ke^{ax}$$ if and only if $$ax=kae^{ax}.$$
Let $y=ax$. Then this is true of and only if $$y=kae^y,$$ that is, $e^y=my$ where $m=\frac{1}{ka}$. It's easy to characterise when this has 0, 1 or 2 solutions, and that gives you the values of $k$ and $a$ which give rise to 0, 1 or 2 solutions.
If $a\neq b$, then the cases $k_1=0$ or $k_2=0$ have been covered above.
Assume for now that $k_1,k_2>0$, and let $x=y+q$. Then $$x=k_1e^{ax}-k_2e^{bx}$$ if and only if $$y+q=k_1e^{ay}e^{aq}+k_2e^{by}e^{bq}\\=k_1e^{aq}\left(e^{ay}+\frac{k_2}{k_1}e^{(b-a)q}e^{by}\right).$$ Letting $$q=\frac{\ln(k_1/k_2)}{b-a},$$ this becomes
$$\frac{y+q}{k_1e^{aq}}=e^{ay}-e^{by}.$$
Then, letting $z=\frac{y}{b-a}$, this becomes $$mz+c=e^{dz}\left(e^z-1\right),$$ where $m$, $c$ and $d$ are expressions in $a$, $b$, $q$, $k_1$ and so on.
Now, it's a matter of characterising the tangent lines to $e^{dz}\left(e^z-1\right)$, and translating those characterisations into regions in $(k_1,k_2,a,b)$ space. This almost-one-parameter problem is much more manageable than the original four-parameter problem.
If $k_1<0$ and $k_2<0$ you can proceed similarly to the above. If $k_1$ and $k_2$ have opposite sign, you can again proceed as above, but you'll need to characterise tangents to $e^{dz}\left(e^z+1\right)$
A: Here is a complete solution for the case $b > a \color{red}{>0}$.
(The methodology used here is easily adaptable to the cases where either $a$ or $b$ or both are negative.)
Let us transform the initial equation:
$$x=k_0e^{ax}-k_1e^{bx}$$
Multiplying its LHS and RHS by $\frac{1}{k_1}e^{-bx}$ gives:
$$\frac{1}{k_1}xe^{-bx}=\frac{k_0}{k_1}e^{(a-b)x}-1$$
Using the change of variable $X=\frac{1}{k_1}x \iff x=k_1X$, we obtain:
$$Xe^{-cX}=ke^{dX}-1\tag{1}$$
with
$$c:=bk_1, \ \ k:=\dfrac{k_0}{k_1},  \ 
 d:=(a-b)k_1\tag{2}$$
We will rename variable $X$ into $x$ because it is more usual:
$$xe^{-cx}=ke^{dx}-1 \tag{3}$$
(keeping equation (1) when we will have to retrieve the true values of the roots).
The roots of equation (3) can be considered as the abscissas of intersection points of the 2 curves with equations
$$f_1(x)=y=xe^{-cx} \ \ \text{and} \ \ f_2(x)=y=ke^{dx}-1$$
($f_1$ is first increasing, passing through $0$, has a maximum, then decreases asymptoticaly to $0$ ; $f_2$ increasing from $y=-1_+$ to $y=+\infty$)
Using the (free) Desmos tool, it is possible by playing on the sliders (you are invited to do it...) to appreciate the different cases, with $0$ or $2$ roots like in the example here:

(the cases where there is a single root can be considered as a double root, therefore can be omitted as a limit case of coalescence of 2 roots).
The unique root $x^*$ of equation $f_2(x)=0$ (the place where its curve crosses the $x$ axis) can be computed easily:
$$x^*=-\frac{1}{c} \ln k=-\dfrac{1}{bk_1}\ln \dfrac{k_1}{k_0}$$
(taking (2) into account).
a) If $x^*>0$ there are certainly two roots, one negative, one positive.
b) If $x^*<0$ there are no roots.
c) If $x^*=0$ which happens iff $k=1$: either

*

*$c<1$ : $2$ roots, one root $0$, the other $>0$;


*$c>1$ : $2$ roots, one root $0$, the other $<0$;


*$c=1$ : $1$ (double) root $0$.
(in fact, these distinctions come from the respective values of the slopes of $f_1$ and $f_2$ in $x=0$).
