Solving the exponential equation $x^2 = e^{-mx}\cdot k$ I just had this problem come up at work, as  part of a simulation where I had to solve the equation mentioned above (where $m$ and $k$ are constants). I googled solving exponential equations and I got so far as realizing that I need to log both sides of the equation resulting in:
$$2\ln x = -mx + \ln k$$
The above form of the equation seems more intractable than the first and am at a loss regarding how to proceed. Can someone please give me a hint as to the way forward?
 A: When you have exponential and linear or quadratic function in same equation, you must use Lambert-$\operatorname{W}$ function which is defined as inverse function of $f(x)=e^xx$.
$$2\ln x=-mx+\ln k$$
$$x^2=e^{-mx}k$$
$$e^{-mx}k=x^2$$
$$e^{-mx}x^{-2}=\dfrac1k$$
$$e^{\dfrac{mx}2}x=\sqrt k$$
$$e^{\dfrac{mx}2}\dfrac{mx}2=\dfrac{m\sqrt k}2$$
$$\dfrac{mx}2=\operatorname{W}_k\left(\dfrac{m\sqrt k}2\right),k\in\mathbb{Z}$$
$$x=\dfrac{2\operatorname{W}_k\left(\dfrac{m\sqrt k}2\right)}{m}$$
A: You can assume $k>0$ as the equation, for $k<0$, has no solution and it is trivial for $k=0$.
Set $x=t\sqrt{k}$ and $q=m\sqrt{k}$. Then the equation becomes
$$
t^2=e^{-qt}
$$
We may also assume $q>0$ (that is, $m>0$), because otherwise just changing $t$ into $-t$ would bring us into the same form. The case $m=0$ is again trivial.
Taking logarithms, the equation becomes $2\log|t|=-qt$, so we can consider the function
$$
f(t)=2\log|t|+qt.
$$
Note that $t=0$ is not a solution of the equation $t^2=e^{-qt}$, so discarding it from the domain of $f$ is not a problem.
We have:
$$
\lim_{t\to-\infty}f(t)=-\infty,
\qquad
\lim_{t\to0}f(t)=-\infty,
\qquad
\lim_{t\to\infty}f(t)=\infty.
$$
Let's consider
$$
f'(t)=\frac{2}{t}+q=\frac{2+qt}{t}
$$
that is positive for $t<-q/2$ or $t>0$. Because of the limits computed above, we see that a solution always exist in the interval $(0,\infty)$.
In order to find possible negative solutions, we need to see whether the maximum of $f$ in the interval $(-\infty,0)$, which is attained at $-2/q$, is positive. Now
$$
f(-2/q)=2\log\frac{2}{q}-2=2\log\frac{2}{qe}
$$
which is positive for $2>qe$, that is, $0<q<2/e$.
Thus the equation has


*

*three solutions for $0<q<2/e$ (two negative, one positive),

*two solutions for $q=2/e$ (one negative, $-2/q$, one positive)

*one solution for $q>2/e$ (positive).

