The function f ( x ) = log ( 1 + x 1 - x ) satisfies the equation

Question

The function $f (x) = \log (\frac{1 + x}{1 - x})$ satisfies the equation

Smart Achievers · Accepted Answer

(3)

$f (x) = \log (\frac{1 + x}{1 - x})$

or $f (y) = \log (\frac{1 + y}{1 - y})$ (by changing $x$ to $y$ )

Taking $f (x) + f (y)$ ,

$= \log (\frac{1 + x}{1 - x}) + \log (\frac{1 + y}{1 - y}) = \log ((\frac{1 + x}{1 - x}) \cdot (\frac{1 + y}{1 - y}))$

$= \log (\frac{1 + x y + x + y}{1 - y - x + x y}) = \log (\frac{(1 + x y) + (x + y)}{(1 + x y) - (x + y)})$

$= \log (\frac{1 + \frac{x + y}{1 + x y}}{1 - \frac{x + y}{1 + x y}}) = f (\frac{x + y}{1 + x y})$

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