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

Options

A. `f(x+2)-2cd(x+1)+f(x)=0`
B. `f(x+1)+f(x)=f(x(x+1))`
C. `f(x_(1)(x_(2))=f(x_(1)+x_(2))`
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 determine which of the given equations is satisfied by the function \( f(x) = \log_{10} \left( \frac{1+x}{1-x} \right) \). We will evaluate each equation one by one.

### Step 1: Evaluate the first equation
The first equation is:
\[ -2f(x) + 1 + f(x) = 0 \]
This simplifies to:
\[ -f(x) + 1 = 0 \implies f(x) = 1 \]
Now substituting \( f(x) \):
\[ \log_{10} \left( \frac{1+x}{1-x} \right) = 1 \]
This implies:
\[ \frac{1+x}{1-x} = 10 \implies 1+x = 10(1-x) \implies 1+x = 10 - 10x \implies 11x = 9 \implies x = \frac{9}{11} \]
Thus, \( f(x) \) does not equal 1 for all \( x \), so the first equation does not hold for all \( x \).

### Step 2: Evaluate the second equation
The second equation is:
\[ f(x) + 1 + f(x) = f(x) \cdot (x + 1) \]
This simplifies to:
\[ 2f(x) + 1 = f(x) \cdot (x + 1) \]
Substituting \( f(x) \):
\[ 2\log_{10} \left( \frac{1+x}{1-x} \right) + 1 = \log_{10} \left( \frac{1+x}{1-x} \right) \cdot (x + 1) \]
This can be rewritten as:
\[ \log_{10} \left( \left( \frac{1+x}{1-x} \right)^2 \cdot 10 \right) = \log_{10} \left( \frac{1+x}{1-x} \cdot (x + 1) \right) \]
Taking the antilogarithm:
\[ \frac{(1+x)^2}{(1-x)^2} \cdot 10 = \frac{(1+x)(x+1)}{1-x} \]
This does not simplify to an identity, so the second equation also fails.

### Step 3: Evaluate the third equation
The third equation is:
\[ f(x_1) + f(x_2) = f(x_1 x_2) \]
Substituting:
\[ \log_{10} \left( \frac{1+x_1}{1-x_1} \right) + \log_{10} \left( \frac{1+x_2}{1-x_2} \right) = f(x_1) + f(x_2) \]
This simplifies to:
\[ \log_{10} \left( \frac{(1+x_1)(1+x_2)}{(1-x_1)(1-x_2)} \right) = \log_{10} \left( \frac{1+x_1 x_2}{1-x_1 x_2} \right) \]
This does not hold true, so the third equation also fails.

### Step 4: Evaluate the fourth equation
The fourth equation is:
\[ f(x_1) + f(x_2) = f\left( \frac{x_1 + x_2}{1 + x_1 x_2} \right) \]
Substituting:
\[ \log_{10} \left( \frac{1+x_1}{1-x_1} \right) + \log_{10} \left( \frac{1+x_2}{1-x_2} \right) = \log_{10} \left( \frac{1 + \frac{x_1 + x_2}{1 + x_1 x_2}}{1 - \frac{x_1 + x_2}{1 + x_1 x_2}} \right) \]
This simplifies to:
\[ \log_{10} \left( \frac{(1+x_1)(1+x_2)}{(1-x_1)(1-x_2)} \right) = \log_{10} \left( \frac{(1+x_1 + x_2 + x_1 x_2)}{(1-x_1 - x_2 + x_1 x_2)} \right) \]
Both sides are equal, confirming that the fourth equation holds.

### Conclusion
The function \( f(x) = \log_{10} \left( \frac{1+x}{1-x} \right) \) satisfies the fourth equation.