Show that f(2x/(1 + x^2)) = 2f(x) where f(x) = log((1 + x)/(1 - x)).

Steps to Solve

1. Substituting the argument in the function

Start by substituting ( x ) with ( \frac{2x}{1 + x^2} ) in the function ( f(x) ) defined as:

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

Thus, we get

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

2. Simplifying the expression inside the logarithm

To simplify ( \frac{1 + \frac{2x}{1 + x^2}}{1 - \frac{2x}{1 + x^2}} ), perform the operations in the numerator and denominator:

• Numerator:

$$ 1 + \frac{2x}{1+x^2} = \frac{(1+x^2) + 2x}{1+x^2} = \frac{1 + x^2 + 2x}{1 + x^2} $$

• Denominator:

$$ 1 - \frac{2x}{1+x^2} = \frac{(1+x^2) - 2x}{1+x^2} = \frac{1 + x^2 - 2x}{1 + x^2} $$

Combining these gives:

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

3. Recognizing the relation to double the function value

Now recognize that:

$$ 1 + x^2 + 2x = (1 + x)^2 $$
$$ 1 + x^2 - 2x = (1 - x)^2 $$

Thus, we can rewrite the expression as:

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

4. Using logarithm properties

Applying the property of logarithms that states ( \log\left(\frac{a^b}{c^b}\right) = b(\log(a) - \log(c)) ):

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

5. Final step: Proving the equation

Since we have:

$$ f\left(\frac{2x}{1+x^2}\right) = 2f(x) $$

Thus, we have shown that ( f\left(\frac{2x}{1+x^2}\right) = 2f(x) ).