Consider $f(x) = e^x x^n - (1-cx)$ where $n \ge 1$.
$f(0) = -1$.
$f(1) = e - 1 +c $,
so if $c > 1-e$
there is a root between $0$ and $1$.
If $ c \le 1-e$,
$f(x) = e^x x^n - 1 -|c|x$,
so, since $e^x > 1+x$ for $x > 0$,
if $x_0 = \max(1, |c|^{1/(n-1)})$,
$x_0^{n-1}\ge |c|$ so
$x_0^n \ge c x_0$
and $f(x_0) > 0$
so there is a root between $0$ and $x_0$.
In either case, we know an interval where there is a positive root.
If $c > 0$, then
$f'(x) = e^x x^{n-1}(x+n) + c > 0$,
so $f$ is increasing and there can be only one positive root and almost any root finder will do.
If $c \le 0$, we might have to do more, but it is late and I am tired.