Answer to Question #204749 in Calculus for Anuj

Question

Prove that

x<log(1/1-x)< x/1-x, 0<x<1

Answer 1

Solution:

Let $f(x)=\log (\frac1{1-x})$ in [0, 1]

Since f(x) satisfies the condition of L.M.V. theorem in [0, 1], there exists $\theta(0<\theta<1)$ such that

$\dfrac{f(x)-f(0)}{x-0}=f^{\prime}(\theta x) \\ \Rightarrow \quad \dfrac{\log (\frac1{1-x})}{x}=\dfrac{}{}\dfrac{1}{1-\theta x}$

$\Rightarrow \quad \log (\frac1{1-x})=\dfrac{x}{1-\theta x} \quad \ldots (i)$

Now, $\quad 0<\theta<1, 0<x<1 \Rightarrow \theta x<x$

$\Rightarrow \quad -\theta x>-x \\\Rightarrow \quad 1-\theta x>1-x \\ \Rightarrow \quad \frac{1}{1-\theta x}<\frac{1}{1-x} \\\Rightarrow \quad \frac{x}{1-\theta x}<\frac{x}{1-x} \quad \ldots (ii)$

Again $\quad 0<\theta<1, 0<x<1$

$\Rightarrow \quad \theta x>0 \\\Rightarrow \quad -\theta x<0 \\ \Rightarrow \quad 1-\theta x<1 \\ \Rightarrow \quad \frac{1}{1-\theta x}>1 \\ \Rightarrow \quad \frac{x}{1-\theta x}>x \quad \ldots (iii)$

From (i), (ii) and (iii), we get, $x< \log (\frac1{1-x})<\dfrac{x}{1-x}, 0<x<1$