Prove the inequality Using Mean Value theorem Show that : $$x < \log\Bigl(\frac{1}{1-x}\Bigr) < \frac{x}{1 - x}\,;$$
If $$0 < x < 1$$
Solution:
If $$f(x) = \log\Bigl(\frac{1}{1 - x}\Bigr)$$
Then the function is continuous in [0, x] 
And also differentiable in (0, x) 
So We can apply Lagrange's MVT on f(x). 
Fine! 
So $$f'(x) =\frac{1}{1 - x}$$
As per LMVT : $$f(x) - f(0) = xf'(θx) ;$$
Where : $$0 < θ < 1$$
=> $$xf'(θx) = - \log(1 - x) $$
=> $$\frac{x} {1 - θx} = - \log(1 - x)$$
=>  $$\frac{x} {1 - θx} = - \log(1 - x)......... (1)$$
So, $$0 < θx < x$$
Or  $$0 > - θx >  - x $$
Or  $$1 >  1 - θx  >  1 - x $$
Or  $$1 < \frac{1} {1 - θx} < \frac{1} {1 - x} $$ 
Or  $$x <  \frac{x} {1 - θx} <\frac{x}{1 - x} $$
Or  $$x <  - \log(1 - x) < \frac{x} {1 - x}... \text{ from } (1)$$
Or
$$x < \log\Bigl(\frac{1}{1 - x}\Bigr) <\frac{x}{1 - x}$$

Hint:
The inequalities should be reversed.
Start from the well-known inequality
$$\log u\le u-1\quad\text{for all }u>0\qquad(< \text{ if } u\ne1),$$
and make the relevant substitutions.
Some details:
First set $\;u=\dfrac1{1-x}$ (which is positive since $0<x<1$). You obtain instantly
$$\log\frac1{1-x} <\frac1{1-x}-1=\frac x{1-x}.$$
Now set $u=1-x$: you get
$$\log(1-x)<(1-x)-1=-x\iff \log\frac1{1-x}>-(-x)=x.$$
