lower bound for $log(1-x)$ for 0≤$x$≤1-$e^{-1}$


How to show $log(1-x) geq -frac{e}{e-1}x$ for 0≤$x$≤1-$e^{-1}$?
I thought about $log(1+x)geq frac{x}{1+x}$, but don’t know how to apply it. Thanks!