Question

If f(x) = log x, show that f(a + b) = f(ab).​

Answer 1

Answer:

f(a) = log( 1−a 1+a ) , f(b) = log( 1−b 1+b ) f( a+b 1+ab ) = log( 1− a+b 1+ab 1+ a+b 1+ab ) ⇒ f( a+b 1+ab ) = log( 1+ab−a−b 1+ab 1+ab+a+b 1+ab ) = log( (1−a)+b(a−1) (1+a)+b(a+1) ) = log( (1−a)(1−b) (1+a)(1+b) ) = log( 1−a 1+a )+log( 1−b 1+b ) (∵ log ab = log a + log b) = f(a) + f(b).

Answer 2

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).

Step-by-step explanation:

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