Question

f(x) = log{(1-x)/(1+x)} then f{(a+b) /(1+ab)} is

Answer 1

Given,
$f(x) = log( \frac{1 - x}{1 + x} )$
Using the Logarithm property,
$log( \frac{m}{n} ) = log(m) - log(n)$
So,
$f(x) = log(1 - x) - log(1 + x)$
As per the problem,
Here,
$x = \frac{a + b}{1 + ab}$
$f(\frac{a + b}{1 + ab}) = log(1 - \frac{a + b}{1 + ab} ) - log(1 + \frac{ a+b }{1 + ab} )$
$f(\frac{a + b}{1 + ab}) = log(\frac{1 + ab - a - b}{1 + ab} ) - log( \frac{ 1 + ab+ a+b }{1 + ab} )$
$f(\frac{a + b}{1 + ab}) = log(\frac{(1 - a) (1 - b)}{1 + ab} ) - log( \frac{ (1 + a)(1 + b) }{1 + ab} )$
Again using property of logarithm as mentioned above,
$f(\frac{a + b}{1 + ab}) = log(\frac{(1 - a) (1 - b)}{(1 + a ) (1 + b )})$

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