Solving a special case exponential function I am currently solving an inequality for a personal project and have run into a quite specific problem that I am unsure how to solve, I therefore asked wolfram alpha, which solved it using something called the analytical continuation of the log production function, however it does not explain how it goes about solving this. The problem is as follows:
$$M_S \cdot e^{\theta \cdot \Delta t \cdot M_S}=\frac{1}{e^\Phi}$$
Subject to $$\theta>0, \Delta t>0, M_S>0$$
And then solve for $M_S$. Any help in how to solve this is much appreciated, moreover, would the result be identical if it was an inequality i.e.
$$M_S \cdot e^{\theta \cdot \Delta t \cdot M_S}<\frac{1}{e^\Phi}$$
My initial guess would be yes as all values are positive except for $\Phi$ however $e^\Phi$ is strictly positive.
Any and all help is appreciated as I am a little puzzled as how to solve this.
 A: Multiplying both sides by $\theta\,\Delta t$, the equation rewrites as
$$ \theta\,\Delta t\, M_S \,\text{e}^{\theta\,\Delta t\, M_S} = \frac{\theta\,\Delta t}{\text{e}^{\Phi}} \, , $$
which is of the form $X \text{e}^X = Y$, where $X=\theta\,\Delta t\, M_S$ and $Y = \theta\,\Delta t\,\text{e}^{-\Phi}$. The solutions are $X=W(Y)$ where $W$ denotes the Lambert W-function. Here, only one real solution is possible since $Y>0$:
$$ M_S = \frac{1}{\theta\,\Delta t} W\!\left(\frac{\theta\,\Delta t}{\text{e}^{\Phi}}\right) . $$
For $X>0$, the function $X\mapsto X \text{e}^X$ is increasing and positive. Thus, the inequality $X \text{e}^X<Y$ is satisfied for all $X$ such that $0<X<W(Y)$. Here, one has,
$$ 0 < M_S < \frac{1}{\theta\,\Delta t} W\!\left(\frac{\theta\,\Delta t}{\text{e}^{\Phi}}\right) . $$

If you prefer, let us plot the function $X\mapsto X \text{e}^X$ and let us find $X$ such that $X \text{e}^X = Y$ for some $Y > 0$. There is only one real solution $X$, which we denote $X = W(Y) > 0$. Now, let us solve $X \text{e}^X < Y$ for some $Y>0$. Graphically, we see that the inequality holds for all $X < W(Y)$.
