# 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!