How can I reduce the following inequality on $x \in [0, 1]$, and $\epsilon > 0$: $$nx(1-x)^n \leq \epsilon$$ to the this form $$n > N(x, \epsilon)\quad?$$ We can assume $n = 1, 2, 3, \dots$.
My attempt:
$$\ln\big(nx(1-x)^n\big) \leq \ln \epsilon$$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \implies \ln n + n\ln(1-x) \leq \ln\frac{\epsilon}{x}$,
and I am stuck. Since $\ln(1-x) < 0$, I think I must divide by it to change the direction of inequality. But I don't know how to simplify/group $n$ and $\ln n$ while decoupling them from $x$ term.
Ultimately, I want to prove that $\forall n \geq N(x, \epsilon), \quad nx(1-x)^n \leq \epsilon$. Perhaps, I do not need to solve for $n$?