If `log(x^(2) - 16) le log_(e) (4 x -11)`, then

Options

A. `4 lt x le 5 `
B. `x lt - 4 or x gt 4 `
C. `-1 le x le 5 `
D. `x lt -1 or x gt 5 `

Answer

To solve the inequality \( \log(x^2 - 16) \leq \log_e(4x - 11) \), we will follow these steps:

### Step 1: Understand the logarithmic properties
Since logarithms are only defined for positive arguments, we need to ensure both \( x^2 - 16 > 0 \) and \( 4x - 11 > 0 \).

### Step 2: Solve the inequalities for the logarithmic arguments
1. **For \( x^2 - 16 > 0 \)**:
\[ x^2 - 16 = (x - 4)(x + 4) > 0 \]
The critical points are \( x = -4 \) and \( x = 4 \). We analyze the intervals:
- For \( x < -4 \): both factors are negative, so the product is positive.
- For \( -4 < x < 4 \): one factor is negative and one is positive, so the product is negative.
- For \( x > 4 \): both factors are positive, so the product is positive.

Thus, the solution for \( x^2 - 16 > 0 \) is:
\[ x \in (-\infty, -4) \cup (4, \infty) \]

2. **For \( 4x - 11 > 0 \)**:
\[ 4x - 11 > 0 \implies x > \frac{11}{4} \]

### Step 3: Combine the conditions
Now we need to combine the intervals obtained from both inequalities:
- From \( x^2 - 16 > 0 \): \( (-\infty, -4) \cup (4, \infty) \)
- From \( 4x - 11 > 0 \): \( \left( \frac{11}{4}, \infty \right) \)

The intersection of these intervals gives us:
- For \( (-\infty, -4) \) and \( \left( \frac{11}{4}, \infty \right) \): no intersection.
- For \( (4, \infty) \) and \( \left( \frac{11}{4}, \infty \right) \): the intersection is \( (4, \infty) \).

### Step 4: Solve the logarithmic inequality
Next, we solve the inequality:
\[ \log(x^2 - 16) \leq \log_e(4x - 11) \]
This implies:
\[ \frac{x^2 - 16}{4x - 11} \leq 1 \]
Rearranging gives:
\[ x^2 - 16 \leq 4x - 11 \]
\[ x^2 - 4x - 5 \leq 0 \]
Factoring the quadratic:
\[ (x - 5)(x + 1) \leq 0 \]
The critical points are \( x = -1 \) and \( x = 5 \). Analyzing the intervals:
- For \( x < -1 \): both factors are negative, so the product is positive.
- For \( -1 < x < 5 \): one factor is negative and one is positive, so the product is negative.
- For \( x > 5 \): both factors are positive, so the product is positive.

Thus, the solution for \( (x - 5)(x + 1) \leq 0 \) is:
\[ x \in [-1, 5] \]

### Step 5: Combine with previous conditions
Now, we need to find the intersection of \( (4, \infty) \) and \( [-1, 5] \):
- The intersection is \( (4, 5] \).

### Final Answer
Thus, the solution to the inequality \( \log(x^2 - 16) \leq \log_e(4x - 11) \) is:
\[ x \in (4, 5] \]