Prove that
x< log(1/1-x)< x/1-x ; 0<x<1
"\\bigstar" To show
x< log(1/1-x)< x/1-x ; 0<x<1
We can prove this by using
"\\bigstar" Mean Value Theorem
for all x>0
"\\bull" using the mean value theorem
"\\boxed{x< log({1\\over1-x})}"
Here x can't equal to 1
Let
"\\boxed{f(x)=x - log({1\\over1-x})}" ..............1
Since
f(0) = 0
"\\boxed{f'(x) = 1+{1\\over(1-x})}" .............2
for all x>0
f(x)<0
Now
"\\boxed{log({1\\over1-x})<{ x\\over1-x }}" ............3
let
"\\boxed{f(x) = log({1\\over1-x}) - {x\\over1-x }}" ............4
"\\bigstar" according to the mean value theorem, an x0
between a and x
with
"\\boxed{f'(x_o ) = { f(x)-f(a)\\over x-a}}" ...........5
"\\bull" Using the property
, we can prove the inequality
By using (1) ,(2) ,(3), (4) ,(5)
We got
"\\bigstar"
"\\boxed{\nx< log({1\\over1-x})< {x\\over1-x }; 0<x<1}"
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