The function $f(x)=\log\left(\frac{1+x}{1-x}\right)$ log satisfies the equation
Answer & explanation
Correct answer: option 4
$f(x_1)+f(x_2)=\log\frac{1+x_1}{1-x_1}+\log\frac{1+x_2}{1-x_2}=f\left(\frac{x_1+x_2}{1+x_1x_2}\right)$
The function $f(x)=\log\left(\frac{1+x}{1-x}\right)$ log satisfies the equation
Correct answer: option 4
$f(x_1)+f(x_2)=\log\frac{1+x_1}{1-x_1}+\log\frac{1+x_2}{1-x_2}=f\left(\frac{x_1+x_2}{1+x_1x_2}\right)$