Question

If f(x)=log (1+x)/(1-x) then prove that f(2x/1+x²)=2f(x)

Answer 1

please check , It might help.

Answer 2

Question :-

$\sf \: if \: f(x) = log( \frac{ 1 + x}{1 - x} ) \: \: then \: prove \: that \\ \\ \sf f \bigg( \frac{2x}{1 + {x}^{2} } \bigg) = 2f(x)$

Proof :-

Formula used :-

→ a² + b² + 2ab = (a+b)²

→ a² + b² - 2ab = (a - b)²

Explanation :-

Taking LHS

$\to \sf f \bigg(\frac{2x}{1 + {x}^{2} } \bigg) \\ \\ hence \\ \\ \to \sf \log \bigg( \frac{1 + \frac{2x}{1 + {x}^{2} } }{1 - \frac{2x}{1 + {x}^{2} } } \bigg) \\ \\ \to \: \sf \log \bigg( \frac{ \frac{1 + {x}^{2} + 2x }{1 + {x}^{2} } }{ \frac{1 + {x}^{2} - 2x}{1 + {x}^{2} } } \bigg) \\ \\ \to \sf \: \log \bigg( \frac{ {x}^{2} + {1}^{2} + 2 \times 1 \times x }{ {x}^{2} + {1}^{2} - 2 \times 1 \times x} \bigg) \\ \\ \to \sf \log \bigg( \frac{ {(1 + x)}^{2} }{ {(1 - x)}^{2} } \bigg) \\ \\ \to \sf \: log{ \bigg( \frac{1 + x}{1 - x} \bigg) }^{2} \\ \\ \because \bf \: log( {m}^{n} ) = n \log m \\ \\ \to \sf \: 2 log \bigg( \frac{1 + x}{1 - x} \bigg) \\ \\ \to \sf \: 2f(x)$

LHS = RHS

hence proved .