The function `f(x)=log_(10)((1+x)/(1-x))` satisfies the equation

Options

A. `f(x+2)-2f(x+1)+f(x)=0`
B. `f(x_(1))f(x_(2))=f(x_(1)+x_(2))`
C. `f(x_(1))+f(x_(2))=f(x_(1)+x_(3))/(1+x_(1)x_(3))`
D. `f(x_(1))+f(x_(2))=f(x_(1)+x_(2))/(1+x_(1)x_(2))`

Answer

To solve the problem, we need to show that the function \( f(x) = \log_{10} \left( \frac{1+x}{1-x} \right) \) satisfies a certain equation. The equation we want to prove is:

\[ f(x_1 + x_2) = f(x_1) + f(x_2) \]

for \( x_1 \) and \( x_2 \) such that \( x_1 + x_2 < 1 \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \log_{10} \left( \frac{1+x}{1-x} \right) \]

2. **Calculate \( f(x_1 + x_2) \)**:
We need to find \( f(x_1 + x_2) \):
\[ f(x_1 + x_2) = \log_{10} \left( \frac{1 + (x_1 + x_2)}{1 - (x_1 + x_2)} \right) \]
This simplifies to:
\[ f(x_1 + x_2) = \log_{10} \left( \frac{1 + x_1 + x_2}{1 - x_1 - x_2} \right) \]

3. **Calculate \( f(x_1) + f(x_2) \)**:
Now calculate \( f(x_1) \) and \( f(x_2) \):
\[ f(x_1) = \log_{10} \left( \frac{1+x_1}{1-x_1} \right) \]
\[ f(x_2) = \log_{10} \left( \frac{1+x_2}{1-x_2} \right) \]
Adding these:
\[ f(x_1) + f(x_2) = \log_{10} \left( \frac{1+x_1}{1-x_1} \right) + \log_{10} \left( \frac{1+x_2}{1-x_2} \right) \]
Using the property of logarithms \( \log(a) + \log(b) = \log(ab) \):
\[ f(x_1) + f(x_2) = \log_{10} \left( \frac{1+x_1}{1-x_1} \cdot \frac{1+x_2}{1-x_2} \right) \]

4. **Combine the fractions**:
Now we can combine the fractions:
\[ f(x_1) + f(x_2) = \log_{10} \left( \frac{(1+x_1)(1+x_2)}{(1-x_1)(1-x_2)} \right) \]

5. **Expand the numerator and denominator**:
Expanding both:
- Numerator: \( (1+x_1)(1+x_2) = 1 + x_1 + x_2 + x_1 x_2 \)
- Denominator: \( (1-x_1)(1-x_2) = 1 - x_1 - x_2 + x_1 x_2 \)

So we have:
\[ f(x_1) + f(x_2) = \log_{10} \left( \frac{1 + x_1 + x_2 + x_1 x_2}{1 - x_1 - x_2 + x_1 x_2} \right) \]

6. **Conclusion**:
Since both expressions for \( f(x_1 + x_2) \) and \( f(x_1) + f(x_2) \) are equal, we conclude that:
\[ f(x_1 + x_2) = f(x_1) + f(x_2) \]

### Final Result:
Thus, the function \( f(x) = \log_{10} \left( \frac{1+x}{1-x} \right) \) satisfies the equation \( f(x_1 + x_2) = f(x_1) + f(x_2) \).