Lower bound on $k$ when $(1-\varepsilon )^k < \frac{3}{4n}$

Question

I try to found Lower bound on $k$ when $(1-\varepsilon )^k < \frac{3}{4n}$. The lower bound should be $k=\Omega(\log{n})$ but I can't find a way to prove that... I tried to use Bernoulli's inequality but the result was not good enough I also tried to proved using combinatorics methods but that also failed. Anyone got an idea?

Thanks

Edit: $\varepsilon$ is constant s.t. $0<\varepsilon<1$