Question

if f of X is equals to log 1 + X by 1 - X then f of 2A by 1 + a square is equals to​

Answer 1

Step-by-step explanation:

f(x) = log(1+x/1-x)

f(x) = log(1+x) - log(1-x)

To find the given function replace x with (2x/1+x^2).

f(2A/1+A^2) =  log(1+(2x/1+x^2)) - log(1-(2x/1+x^2))  

f(2A/1+A^2) = log((x^2+1+2x)/(1+x^2))-log((x^2+1-2x)/(1+x^2))

On simplifying the above equation we get,

f(2A/1+A^2) = log((x+1)^2/(x-1)^2).

This can be written as 2*log((x+1)/(x-1)).

Hope this helps :)