If [f(x) = _e( 1 - x 1 + x )], [| x |

Question

If $$f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$, $$\left| x \right| < 1$$ then $$f\left( {{{2x} \over {1 + {x^2}}}} \right)$$ is equal to

Accepted Answer

Given, $$f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$

$$f\left( {{{2x} \over {1 + {x^2}}}} \right)$$ = $$\ln \left( {{{1 - {{2x} \over {1 + {x^2}}}} \over {1 + {{2x} \over {1 + {x^2}}}}}} \right)$$

= $$\ln \left( {{{{x^2} - 2x + 1} \over {{x^2} + 2x + 1}}} \right)$$

= $$\ln {\left( {{{1 - x} \over {1 + x}}} \right)^2}$$

= $$2\ln \left( {{{1 - x} \over {1 + x}}} \right)$$

= 2f(x)

Solutions · 1