If `"log"_(4)(3x^(2) +11x) gt 1`, then x lies in the interval

Options

A. `(-4, 1//3)`
B. `(-4, 2)`
C. `[-4, 1//3]`
D. `(-oo, -4)cup (1//3, oo)`

Answer

To solve the inequality \( \log_{4}(3x^{2} + 11x) > 1 \), we will follow these steps:

### Step 1: Understand the condition for the logarithm to be defined
The expression inside the logarithm must be greater than zero:
\[ 3x^{2} + 11x > 0 \]

### Step 2: Factor the quadratic expression
We can factor out \( x \):
\[ x(3x + 11) > 0 \]

### Step 3: Find the critical points
Set each factor to zero to find the critical points:
1. \( x = 0 \)
2. \( 3x + 11 = 0 \) which gives \( x = -\frac{11}{3} \)

### Step 4: Determine the intervals
The critical points divide the number line into intervals:
1. \( (-\infty, -\frac{11}{3}) \)
2. \( (-\frac{11}{3}, 0) \)
3. \( (0, \infty) \)

### Step 5: Test the intervals
We will test each interval to see where the product \( x(3x + 11) \) is positive.

- For \( x < -\frac{11}{3} \) (e.g., \( x = -4 \)):
\[ -4(3(-4) + 11) = -4(-12 + 11) = -4(-1) = 4 > 0 \quad \text{(True)} \]

- For \( -\frac{11}{3} < x < 0 \) (e.g., \( x = -1 \)):
\[ -1(3(-1) + 11) = -1(-3 + 11) = -1(8) = -8 < 0 \quad \text{(False)} \]

- For \( x > 0 \) (e.g., \( x = 1 \)):
\[ 1(3(1) + 11) = 1(3 + 11) = 1(14) = 14 > 0 \quad \text{(True)} \]

### Step 6: Combine the results
From the testing, we find:
- The expression \( 3x^{2} + 11x > 0 \) is true for \( x \in (-\infty, -\frac{11}{3}) \cup (0, \infty) \).

### Step 7: Solve the logarithmic inequality
Now, we need to solve:
\[ \log_{4}(3x^{2} + 11x) > 1 \]
This can be rewritten using the property of logarithms:
\[ 3x^{2} + 11x > 4^{1} \quad \Rightarrow \quad 3x^{2} + 11x > 4 \]

### Step 8: Rearrange the inequality
Rearranging gives:
\[ 3x^{2} + 11x - 4 > 0 \]

### Step 9: Factor the quadratic
We can factor or use the quadratic formula to find the roots:
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{-11 \pm \sqrt{11^{2} - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ = \frac{-11 \pm \sqrt{121 + 48}}{6} = \frac{-11 \pm \sqrt{169}}{6} = \frac{-11 \pm 13}{6} \]
This gives us the roots:
1. \( x = \frac{2}{6} = \frac{1}{3} \)
2. \( x = \frac{-24}{6} = -4 \)

### Step 10: Determine the intervals for the quadratic inequality
The roots divide the number line into intervals:
1. \( (-\infty, -4) \)
2. \( (-4, \frac{1}{3}) \)
3. \( (\frac{1}{3}, \infty) \)

### Step 11: Test the intervals for the quadratic inequality
- For \( x < -4 \) (e.g., \( x = -5 \)):
\[ 3(-5)^{2} + 11(-5) - 4 = 75 - 55 - 4 = 16 > 0 \quad \text{(True)} \]

- For \( -4 < x < \frac{1}{3} \) (e.g., \( x = 0 \)):
\[ 3(0)^{2} + 11(0) - 4 = -4 < 0 \quad \text{(False)} \]

- For \( x > \frac{1}{3} \) (e.g., \( x = 1 \)):
\[ 3(1)^{2} + 11(1) - 4 = 3 + 11 - 4 = 10 > 0 \quad \text{(True)} \]

### Step 12: Combine the results
The intervals where \( 3x^{2} + 11x - 4 > 0 \) are:
- \( (-\infty, -4) \)
- \( (\frac{1}{3}, \infty) \)

### Step 13: Find the intersection of both conditions
The final intervals where both conditions are satisfied are:
- From the first condition: \( (-\infty, -\frac{11}{3}) \cup (0, \infty) \)
- From the second condition: \( (-\infty, -4) \cup (\frac{1}{3}, \infty) \)

The intersection gives us:
- \( (-\infty, -4) \)
- \( (\frac{1}{3}, \infty) \)

### Final Answer
Thus, the solution is:
\[ x \in (-\infty, -4) \cup \left(\frac{1}{3}, \infty\right) \]