Solution set of the inequality `(log)_(0. 8)((log)_6(x^2+x)/(x+4))<0` is

Options

A. `(-4,-3)`
B. `(-3,4)uu(8,oo)`
C. `(-3,oo)`
D. `(-4,-3)uu(8,oo)`

Answer

To solve the inequality \( \log_{0.8} \left( \frac{\log_6(x^2+x)}{x+4} \right) < 0 \), we will follow these steps:

### Step 1: Understanding the logarithm base
Since the base of the logarithm is \(0.8\), which is less than 1, we know that:
\[ \log_{0.8}(A) < 0 \quad \text{if and only if} \quad A > 1 \]
Thus, we can rewrite our inequality:
\[ \frac{\log_6(x^2+x)}{x+4} > 1 \]

### Step 2: Rearranging the inequality
We can rearrange the inequality:
\[ \log_6(x^2+x) > x + 4 \]

### Step 3: Exponentiating both sides
To eliminate the logarithm, we exponentiate both sides with base 6:
\[ x^2 + x > 6^{x + 4} \]

### Step 4: Setting up the function
Let’s define a function:
\[ f(x) = x^2 + x - 6^{x + 4} \]
We need to find where \(f(x) > 0\).

### Step 5: Finding critical points
To find the critical points, we can analyze the function \(f(x)\). We can evaluate \(f(x)\) at some key points:

1. **At \(x = -4\)**:
\[ f(-4) = (-4)^2 + (-4) - 6^{0} = 16 - 4 - 1 = 11 > 0 \]

2. **At \(x = -3\)**:
\[ f(-3) = (-3)^2 + (-3) - 6^{1} = 9 - 3 - 6 = 0 \]

3. **At \(x = 8\)**:
\[ f(8) = 8^2 + 8 - 6^{12} \text{ (which is a very large negative number)} \]

### Step 6: Analyzing the sign of \(f(x)\)
Now we need to analyze the intervals determined by our critical points \(x = -4\) and \(x = -3\):
- For \(x < -4\), \(f(x) > 0\)
- For \(-4 < x < -3\), \(f(x) < 0\)
- For \(x = -3\), \(f(x) = 0\)
- For \(x > -3\), \(f(x)\) will eventually become negative as \(6^{x + 4}\) grows faster than \(x^2 + x\).

### Step 7: Conclusion
Thus, the solution set for the inequality \( \log_{0.8} \left( \frac{\log_6(x^2+x)}{x+4} \right) < 0 \) is:
\[ (-4, -3) \cup (8, \infty) \]