Complete set of real values of x for which `log_((2x-3))(x^(2)-5x-6)` is defined is :

Options

A. `((3)/(2),oo)`
B. `(6, oo)`
C. `((3)/(2),6)`
D. `((3)/(2),2) cup (2, oo)`

Answer

To find the complete set of real values of \( x \) for which \( \log_{(2x-3)}(x^2 - 5x - 6) \) is defined, we need to ensure that both the base of the logarithm and the argument of the logarithm satisfy certain conditions.

### Step 1: Conditions for the base of the logarithm
The base of the logarithm \( 2x - 3 \) must satisfy the following conditions:
1. \( 2x - 3 \neq 0 \)
2. \( 2x - 3 \neq 1 \)
3. \( 2x - 3 > 0 \)

#### Solving these conditions:
1. **From \( 2x - 3 \neq 0 \)**:
\[ 2x \neq 3 \implies x \neq \frac{3}{2} \]

2. **From \( 2x - 3 \neq 1 \)**:
\[ 2x \neq 4 \implies x \neq 2 \]

3. **From \( 2x - 3 > 0 \)**:
\[ 2x > 3 \implies x > \frac{3}{2} \]

### Step 2: Conditions for the argument of the logarithm
The argument of the logarithm \( x^2 - 5x - 6 \) must also be greater than 0:
\[ x^2 - 5x - 6 > 0 \]

#### Factorizing the quadratic:
To solve \( x^2 - 5x - 6 > 0 \), we can factor it:
\[ x^2 - 5x - 6 = (x - 6)(x + 1) \]

#### Finding the roots:
The roots are \( x = 6 \) and \( x = -1 \). We will analyze the sign of the expression \( (x - 6)(x + 1) \).

### Step 3: Sign analysis
We will test the intervals determined by the roots \( -1 \) and \( 6 \):
- For \( x < -1 \): Choose \( x = -2 \) → \( (-2 - 6)(-2 + 1) = (-8)(-1) > 0 \)
- For \( -1 < x < 6 \): Choose \( x = 0 \) → \( (0 - 6)(0 + 1) = (-6)(1) < 0 \)
- For \( x > 6 \): Choose \( x = 7 \) → \( (7 - 6)(7 + 1) = (1)(8) > 0 \)

Thus, the intervals where \( x^2 - 5x - 6 > 0 \) are:
\[ (-\infty, -1) \cup (6, \infty) \]

### Step 4: Combining the conditions
Now we need to find the intersection of the intervals from the base and the argument:
1. From the base conditions, we have \( x > \frac{3}{2} \) (which is \( 1.5 \)).
2. From the argument conditions, we have \( (-\infty, -1) \cup (6, \infty) \).

### Step 5: Finding the intersection
- The interval \( (-\infty, -1) \) does not intersect with \( x > \frac{3}{2} \).
- The interval \( (6, \infty) \) does intersect with \( x > \frac{3}{2} \).

Thus, the final solution is:
\[ \text{The complete set of real values of } x \text{ for which } \log_{(2x-3)}(x^2 - 5x - 6) \text{ is defined is } (6, \infty). \]