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

Options

A. `(f(x))^(2)`
B. 1
C. `2f(x)`
D. `f(1-x)/(1+x)`

Answer

To solve the problem, we need to find \( f\left(\frac{2x}{1-x^2}\right) \) 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 expressions**:
Notice that:
\[ 1 + 2x - x^2 = (1+x)^2 \quad \text{and} \quad 1 - 2x - x^2 = (1-x)^2 \]
Therefore, we can rewrite the logarithm:
\[ f\left(\frac{2x}{1-x^2}\right) = \log\left(\frac{(1+x)^2}{(1-x)^2}\right) \]

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

7. **Relate back to the original function**:
Since \( f(x) = \log\left(\frac{1+x}{1-x}\right) \), we have:
\[ f\left(\frac{2x}{1-x^2}\right) = 2f(x) \]

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