The function `f(x)=log_(2x-5)(x^(2)-3x-10)` is defined for all belonging to

Options

A. `[5,oo)`
B. `(5,oo)`
C. `(-oo,+5)`
D. none of these

Answer

To determine the domain of the function \( f(x) = \log_{(2x-5)}(x^2 - 3x - 10) \), we need to ensure two conditions are satisfied:

1. The base of the logarithm, \( 2x - 5 \), must be greater than 0 and not equal to 1.
2. The argument of the logarithm, \( x^2 - 3x - 10 \), must be greater than 0.

### Step 1: Analyze the base of the logarithm

The base is given by:
\[ 2x - 5 > 0 \]
Solving this inequality:
\[ 2x > 5 \implies x > \frac{5}{2} \]

Next, we ensure the base is not equal to 1:
\[ 2x - 5 \neq 1 \]
Solving this:
\[ 2x \neq 6 \implies x \neq 3 \]

### Step 2: Analyze the argument of the logarithm

The argument is given by:
\[ x^2 - 3x - 10 > 0 \]
To solve this inequality, we first factor the quadratic:
\[ x^2 - 3x - 10 = (x - 5)(x + 2) \]
Now we need to find when this product is greater than 0. The roots of the equation are \( x = 5 \) and \( x = -2 \).

### Step 3: Test intervals

We will test the intervals determined by the roots \( -2 \) and \( 5 \):

1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \)
\[ (-3 - 5)(-3 + 2) = (-8)(-1) > 0 \quad \text{(True)} \]

2. **Interval \( (-2, 5) \)**: Choose \( x = 0 \)
\[ (0 - 5)(0 + 2) = (-5)(2) < 0 \quad \text{(False)} \]

3. **Interval \( (5, \infty) \)**: Choose \( x = 6 \)
\[ (6 - 5)(6 + 2) = (1)(8) > 0 \quad \text{(True)} \]

### Step 4: Combine the results

From the analysis:
- From the base condition, we have \( x > \frac{5}{2} \) and \( x \neq 3 \).
- From the argument condition, we have \( x \in (-\infty, -2) \cup (5, \infty) \).

The valid intervals that satisfy both conditions are:
- The interval \( (5, \infty) \) is valid since it satisfies both conditions.
- The interval \( (-\infty, -2) \) is not valid since it does not satisfy \( x > \frac{5}{2} \).

### Final Domain

Thus, the domain of the function \( f(x) \) is:
\[ x \in (5, \infty) \]

### Conclusion

The correct option is **B**: \( (5, \infty) \).