What is the proof that $|\ln (1-x)|\le 2|x|$ if $|x|\le \frac{1}{2}$?

Question

I'm trying to prove that $$|\ln (1-x)|\le 2|x| \, \text{if} \, |x|\le \frac{1}{2}.$$ What I've got so far is $$\begin{align*}e^{x}&=1+x+\dfrac{x^{2}}{2!}+\cdots\\1+x&\le e^{x}\\1-x&\le e^{-x}\\ \ln (1-x)&\le -x.\end{align*}$$ However, I'm stuck here since when I take the absolute value of both sides I get nonsense (well, not exactly, it holds only for $x=0$): $$|\ln (1-x)|\le |x|$$ I don't know how to proceed in order to prove the inequality mentioned in the question.

Edit: Also, why is taking the absolute value of both sides (as it seems) forbidden?

Answer 1

How about $$|\ln(1-x)|=\left|-\sum_{n=1}^\infty\frac{x^n}n\right| \le\sum_{n=1}^\infty\frac{|x|^n}n\le\sum_{n=1}^\infty|x|^n=\frac{|x|}{1-|x|}$$ etc.?

Answer 2

Let $x\in [-1/2,1/2].$ By the MVT we have

$$\ln (1-x) = \ln (1) -\frac{1}{1-c_x}x$$

for some $c_x$ between $0$ and $x.$ Thus

$$|\ln (1-x)| = \left|\frac{1}{1-c_x}x\right|\le 2|x|.$$

We get the factor of $2$ on the right because $|c_x|\le 1/2,$ hence $1/(1-c_x)\le 1/(1-1/2)=2.$