If your signal looks like f(t) = K•u(t)e^at with u(t) = {1 if t≥0, 0 else}:
If Real(a) > 0, then your signal will eventually blow up.
If Real(a) < 0, then you signal will not blow up. In fact, your signal will have a maximum absolute value of |K|, and it will approach zero as time goes on.
If Real(a) = 0, it is either a complex sinusoid or a constant. In either case, it is bounded with maximum absolute value of |K|. It very much does not blow up.
So e pops up all the time in stable systems and bounded signals because the function e^at solves the common differential equation dx/dt = ax(t) with x(0)=1 regardless of the value of a, particularly regardless of whether or not the real part of a causes the solution to blow up.
If your signal looks like f(t) = K•u(t)e^at with u(t) = {1 if t≥0, 0 else}:
So e pops up all the time in stable systems and bounded signals because the function e^at solves the common differential equation dx/dt = ax(t) with x(0)=1 regardless of the value of a, particularly regardless of whether or not the real part of a causes the solution to blow up.