The domain of definition of
` log_4 log_5 log_3 (18 x -x^2 - 77)` is given by

Options

A. `(3,5)`
B. ` (7,11)`
C. ` (8,10)`
D. None of these

Answer

To find the domain of the function \( \log_4 \log_5 \log_3 (18x - x^2 - 77) \), we need to ensure that all logarithmic expressions are defined and positive.

### Step-by-step Solution:

1. **Identify the innermost logarithm**: We start with the innermost function, which is \( \log_3 (18x - x^2 - 77) \). For this logarithm to be defined, the argument must be greater than 0:
\[ 18x - x^2 - 77 > 0 \]

2. **Rearrange the inequality**: Rearranging gives us:
\[ -x^2 + 18x - 77 > 0 \]
or equivalently,
\[ x^2 - 18x + 77 < 0 \]

3. **Find the roots of the quadratic**: We can find the roots of the quadratic equation \( x^2 - 18x + 77 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{18 \pm \sqrt{18^2 - 4 \cdot 1 \cdot 77}}{2 \cdot 1} \]
\[ = \frac{18 \pm \sqrt{324 - 308}}{2} = \frac{18 \pm \sqrt{16}}{2} = \frac{18 \pm 4}{2} \]
This gives us the roots:
\[ x = \frac{22}{2} = 11 \quad \text{and} \quad x = \frac{14}{2} = 7 \]

4. **Determine the intervals**: The quadratic \( x^2 - 18x + 77 \) opens upwards (since the coefficient of \( x^2 \) is positive), so it will be negative between the roots:
\[ 7 < x < 11 \]

5. **Check the next logarithm**: Now we need to consider the next logarithm, \( \log_5 \log_3 (18x - x^2 - 77) \). For this to be defined, we need:
\[ \log_3 (18x - x^2 - 77) > 0 \]
This implies:
\[ 18x - x^2 - 77 > 1 \quad \Rightarrow \quad 18x - x^2 > 78 \]
Rearranging gives:
\[ -x^2 + 18x - 78 > 0 \quad \Rightarrow \quad x^2 - 18x + 78 < 0 \]

6. **Find the roots of the new quadratic**: Using the quadratic formula again:
\[ x = \frac{18 \pm \sqrt{18^2 - 4 \cdot 1 \cdot 78}}{2 \cdot 1} = \frac{18 \pm \sqrt{324 - 312}}{2} = \frac{18 \pm \sqrt{12}}{2} \]
\[ = \frac{18 \pm 2\sqrt{3}}{2} = 9 \pm \sqrt{3} \]

7. **Determine the intervals**: The quadratic \( x^2 - 18x + 78 \) also opens upwards, so it will be negative between the roots:
\[ 9 - \sqrt{3} < x < 9 + \sqrt{3} \]

8. **Combine the intervals**: The domain of the function is the intersection of the two intervals:
- From step 4: \( 7 < x < 11 \)
- From step 7: \( 9 - \sqrt{3} < x < 9 + \sqrt{3} \)

The approximate values are:
- \( 9 - \sqrt{3} \approx 7.268 \)
- \( 9 + \sqrt{3} \approx 10.732 \)

Therefore, the intersection is:
\[ 9 - \sqrt{3} < x < 11 \]

### Final Domain:
The domain of the function is:
\[ (9 - \sqrt{3}, 11) \]