If f(x) = log(1 + x)/(1 - x), show that (2x)/(1 + x^2) = 2f(x).

Steps to Solve

1. Substituting into ( f(x) )

Start with the expression we want to show, $$ f\left(\frac{2x}{1+x^2}\right) = \log\left(\frac{1 + \frac{2x}{1+x^2}}{1 - \frac{2x}{1+x^2}}\right) $$

2. Simplifying the argument of the logarithm

Calculate ( 1 + \frac{2x}{1+x^2} ) and ( 1 - \frac{2x}{1+x^2} ):

For the numerator: $$ 1 + \frac{2x}{1+x^2} = \frac{(1+x^2) + 2x}{1+x^2} = \frac{1 + 2x + x^2}{1 + x^2} $$

For the denominator: $$ 1 - \frac{2x}{1+x^2} = \frac{(1+x^2) - 2x}{1+x^2} = \frac{1 - 2x + x^2}{1 + x^2} $$

3. Combining fractions in the logarithm

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

4. Applying logarithmic properties

Use the properties of logarithms to simplify: $$ \log\left(\frac{1 + 2x + x^2}{1 - 2x + x^2}\right) = \log(1 + 2x + x^2) - \log(1 - 2x + x^2) $$

5. Simplifying further

Notice that: $$ 1 + 2x + x^2 = (1 + x)^2 \quad \text{and} \quad 1 - 2x + x^2 = (1 - x)^2 $$

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

6. Finalizing the expression

Using another logarithmic property, we have: $$ \log\left(\frac{(1+x)^2}{(1-x)^2}\right) = 2\log\left(\frac{1+x}{1-x}\right) $$

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