If f (x) = \(log\big(\frac{1-x}{1+x}\big)\) show that f (a) + f (b) =

1.	If f (x) = \(log\big(\frac{1-x}{1+x}\big)\) show that f (a) + f (b) = \(f\big(\frac{a+b}{1+ab}\big)\) .
Answer» f(a) = \(log\big(\frac{1-a}{1+a}\big)\) , f(b) = \(log\big(\frac{1-b}{1+b}\big)\) \(f\big(\frac{a+b}{1+ab}\big)\) = \(log\big(\frac{1- \frac{a+b}{1+ab}}{1+ \frac{a+b}{1+ab}}\big)\) ⇒ \(f\big(\frac{a+b}{1+ab}\big)\) = \(log\big(\frac{\frac{1+ab-a-b}{1+ab}}{\frac{1+ab+a+b}{1+ab}}\big)\) = \(log\big(\frac{(1-a)+b(a-1)}{(1+a)+b(a+1)}\big)\) = \(log\big(\frac{(1-a)(1-b)}{(1+a)(1+b)}\big)\) = \(log\big(\frac{1-a}{1+a}\big)+log\big(\frac{1-b}{1+b}\big)\) (∵ log ab = log a + log b) = f(a) + f(b).

If f (x) = \(log\big(\frac{1-x}{1+x}\big)\) show that f (a) + f (b) = \(f\big(\frac{a+b}{1+ab}\big)\) .

Answer»

f(a) = \(log\big(\frac{1-a}{1+a}\big)\) , f(b) = \(log\big(\frac{1-b}{1+b}\big)\)

\(f\big(\frac{a+b}{1+ab}\big)\) = \(log\big(\frac{1- \frac{a+b}{1+ab}}{1+ \frac{a+b}{1+ab}}\big)\)

⇒ \(f\big(\frac{a+b}{1+ab}\big)\) = \(log\big(\frac{\frac{1+ab-a-b}{1+ab}}{\frac{1+ab+a+b}{1+ab}}\big)\)

= \(log\big(\frac{(1-a)+b(a-1)}{(1+a)+b(a+1)}\big)\) = \(log\big(\frac{(1-a)(1-b)}{(1+a)(1+b)}\big)\)

= \(log\big(\frac{1-a}{1+a}\big)+log\big(\frac{1-b}{1+b}\big)\) (∵ log ab = log a + log b)

= f(a) + f(b).