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.

  • $\begingroup$ What is the source of the precalculus problem? $\endgroup$ Nov 16 '21 at 3:49
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    $\begingroup$ I came to this problem while working on a math model of cellular membrane transport and chemical reaction dynamics (steady state condition), so there's no source here. I'm not sure which tag I should use and like you probably suggested, this is not really a pre-calculus problem maybe it's more like algebraic geometry kinda problem? I guess it depends on how should I approach this problem. $\endgroup$
    – Aiden Bhe
    Nov 16 '21 at 3:57
  • $\begingroup$ Have you studied Calculus? $\endgroup$ Nov 16 '21 at 3:58
  • $\begingroup$ Yes, I have studied Calculus. $\endgroup$
    – Aiden Bhe
    Nov 16 '21 at 4:01
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    $\begingroup$ I would approach this issue from a pragmatic point of view: make systematic computer simulations for parameter $a$ ranging from $a_0$ to $a_1$, parameter $b$ ranging from $b_0$ to $b_1$, etc. with small steps ... counting each time how many roots there are and placing each result in a 3D space with coordinates $(a',b',k_1)$ with a color code (blue point for 2 roots, red point for one root, etc. (3D space if you apply my proposal to eliminate parameter $k_0$). Then understand where (and if possible why) transition "membranes" (:)) appear. $\endgroup$
    – Jean Marie
    Nov 16 '21 at 5:41

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 $e^{dz}\left(e^z+1\right)$

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    $\begingroup$ In the bottom paragraph, you haven't to consider these cases because it is said in the question that $k_0,k_1$ are $>0$. $\endgroup$
    – Jean Marie
    Nov 16 '21 at 10:45
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    $\begingroup$ Thank you! I think I can continue on from this. Changing of variables save the day again. $\endgroup$
    – Aiden Bhe
    Nov 16 '21 at 15:04
  • $\begingroup$ @JeanMarie thanks, I missed that. It makes things much simpler :) $\endgroup$ Nov 17 '21 at 4:42
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    $\begingroup$ [+1} Obtaining a form $mz+c=e^{dz}\left(e^z-1\right)$ with consideration of tangent lines as frontier cases is indeed a good way to approach this kind of issues. $\endgroup$
    – Jean Marie
    Nov 17 '21 at 9:00

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:


Multiplying its LHS and RHS by $\frac{1}{k_1}e^{-bx}$ gives:


Using the change of variable $X=\frac{1}{k_1}x \iff x=k_1X$, we obtain:



$$c:=bk_1, \ \ k:=\dfrac{k_0}{k_1}, \ d:=(a-b)k_1\tag{2}$$

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:

enter image description 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 $x=0$).

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    $\begingroup$ In fact, I just noticed that I have considered cases where $a>b \color{red}{>0}$ (see the first sentence I just added). Your comments ? $\endgroup$
    – Jean Marie
    Nov 16 '21 at 12:56
  • $\begingroup$ Wow, I really appreciate your very detailed answer. It is much easier to see after the change of variable and the case where $a>b>0$, as you have explained, or $b>0$ in general are probably the easiest ones. Things really start to get interesting when $b<0$ and 3 roots become possible (in both cases where $a>b$ and $b<a$), since there are many way for a pair of roots to vanish. $\endgroup$
    – Aiden Bhe
    Nov 16 '21 at 14:34

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