Show that a given transcendental function is uniquely solvable. I'm looking for a theorem to use to show that the following transcendental function is uniquely solvable. I don't care really about the solution.
$$ \rho = \mathrm{e}^{x-\rho t} $$
How might I show this function is uniquely solvable for $\rho$?
 A: The equation is equivalent to $\rho t e^{\rho t} = te^x$.
Let $y=\rho t$ and $a= te^x$. Then $y e^y = a$.
The equation $y e^y = a$ has a solution iff $a \ge -1/e$. The solution is unique iff $a \ge 0$.
All this follows from plotting a graph. Uniqueness also follows from taking the derivative and seeing that it is positive for $y \ge 0$ when $y e^y \ge 0$.

A: Your proposition is true only if $t$ is non-negative. If $t=0$, this is trivial to see. If $t \gt 0$, transform the equation to
$$\rho e^{\rho t} = e^x$$
The right hand side is positive, so the left hand side must be positive as well, hence $\rho$ is positive. Now, the left hand side is a product of 2 positive values $\rho$ and $e^{\rho t}$, both of which are strictly increasing functions of $\rho$ for positive $\rho$ (and $t$). So the left hand side is a strictly increasing function of $\rho$ and thus the equation has at most one solution. Since the left hand side becomes $0$ for $\rho=0$ and goes to infinity for $\rho \to \infty $, it has exactly one solution.
However, the proposition is not true for negative $t$. The function $f(\rho)=\rho e^{\rho t}$ has a maximum at $\rho=-\frac{1}{t}$, so the above equation will have 2 possible solutions for "small enough $x$". 
