If `f(x) = log ((1+x)/(1-x))`, then `f((2x)/(1+x^(2)))` is equal to

Options

A. `(f(x))^(2)`
B. `(f(x))^(3)`
C. `2f(x)`
D. 3f(x)

Answer

To solve the problem, we need to evaluate \( f\left(\frac{2x}{1+x^2}\right) \) given that \( f(x) = \log\left(\frac{1+x}{1-x}\right) \).

### Step-by-Step Solution:

1. **Substitute the value into the function**:
We start by substituting \( \frac{2x}{1+x^2} \) into the function \( f(x) \).
\[ f\left(\frac{2x}{1+x^2}\right) = \log\left(\frac{1+\frac{2x}{1+x^2}}{1-\frac{2x}{1+x^2}}\right) \]

2. **Simplify the numerator**:
The numerator becomes:
\[ 1 + \frac{2x}{1+x^2} = \frac{(1+x^2) + 2x}{1+x^2} = \frac{1 + 2x + x^2}{1+x^2} \]

3. **Simplify the denominator**:
The denominator becomes:
\[ 1 - \frac{2x}{1+x^2} = \frac{(1+x^2) - 2x}{1+x^2} = \frac{1 - 2x + x^2}{1+x^2} \]

4. **Combine the fractions**:
Now we can combine the numerator and denominator:
\[ f\left(\frac{2x}{1+x^2}\right) = \log\left(\frac{\frac{1 + 2x + x^2}{1+x^2}}{\frac{1 - 2x + x^2}{1+x^2}}\right) = \log\left(\frac{1 + 2x + x^2}{1 - 2x + x^2}\right) \]

5. **Factor the numerator and denominator**:
Notice that:
- The numerator \( 1 + 2x + x^2 \) can be factored as \( (x+1)^2 \).
- The denominator \( 1 - 2x + x^2 \) can be factored as \( (x-1)^2 \).

Thus, we have:
\[ f\left(\frac{2x}{1+x^2}\right) = \log\left(\frac{(x+1)^2}{(x-1)^2}\right) \]

6. **Use properties of logarithms**:
We can use the property of logarithms that states \( \log\left(\frac{a}{b}\right) = \log(a) - \log(b) \):
\[ f\left(\frac{2x}{1+x^2}\right) = \log((x+1)^2) - \log((x-1)^2) \]

This can be simplified further:
\[ = 2\log(x+1) - 2\log(x-1) = 2\left(\log(x+1) - \log(x-1)\right) \]

7. **Express in terms of \( f(x) \)**:
Recall that \( f(x) = \log\left(\frac{1+x}{1-x}\right) \). We can express:
\[ f(x) = \log(x+1) - \log(x-1) \]

Therefore, we can write:
\[ f\left(\frac{2x}{1+x^2}\right) = 2f(x) \]

### Final Result:
Thus, the result is:
\[ f\left(\frac{2x}{1+x^2}\right) = 2f(x) \]