The domain of definition of `f(x)=log_(0.5){-log_(2)((3x-1)/(3x+2))}`, is

Options

A. `(-oo,-1//3)`
B. `(-1//3,oo)`
C. `(1//3,oo)`
D. `[1//3,oo)`

Answer

To find the domain of the function \( f(x) = \log_{0.5} \left( -\log_{2} \left( \frac{3x-1}{3x+2} \right) \right) \), we need to ensure that the arguments of the logarithmic functions are valid.

### Step 1: Determine the condition for the inner logarithm
The inner logarithm \( \log_{2} \left( \frac{3x-1}{3x+2} \right) \) must be positive:
\[ \frac{3x-1}{3x+2} > 0 \]

### Step 2: Analyze the fraction
The fraction \( \frac{3x-1}{3x+2} \) is positive when both the numerator and denominator are either both positive or both negative.

1. **Numerator**: \( 3x - 1 > 0 \) implies \( x > \frac{1}{3} \)
2. **Denominator**: \( 3x + 2 > 0 \) implies \( x > -\frac{2}{3} \)

### Step 3: Find critical points
The critical points from the inequalities are:
- \( x = \frac{1}{3} \) (where the numerator is zero)
- \( x = -\frac{2}{3} \) (where the denominator is zero)

### Step 4: Test intervals
We will test the intervals determined by these critical points:
- \( (-\infty, -\frac{2}{3}) \)
- \( (-\frac{2}{3}, \frac{1}{3}) \)
- \( (\frac{1}{3}, \infty) \)

1. **Interval \( (-\infty, -\frac{2}{3}) \)**: Choose \( x = -1 \)
\[ \frac{3(-1)-1}{3(-1)+2} = \frac{-3-1}{-3+2} = \frac{-4}{-1} = 4 > 0 \]
This interval is valid.

2. **Interval \( (-\frac{2}{3}, \frac{1}{3}) \)**: Choose \( x = 0 \)
\[ \frac{3(0)-1}{3(0)+2} = \frac{-1}{2} < 0 \]
This interval is not valid.

3. **Interval \( (\frac{1}{3}, \infty) \)**: Choose \( x = 1 \)
\[ \frac{3(1)-1}{3(1)+2} = \frac{3-1}{3+2} = \frac{2}{5} > 0 \]
This interval is valid.

### Step 5: Combine results
From the analysis, we find that the valid intervals for \( \frac{3x-1}{3x+2} > 0 \) are:
\[ (-\infty, -\frac{2}{3}) \cup \left(\frac{1}{3}, \infty\right) \]

### Step 6: Condition for the outer logarithm
Next, we need the outer logarithm \( -\log_{2} \left( \frac{3x-1}{3x+2} \right) \) to be positive:
\[ -\log_{2} \left( \frac{3x-1}{3x+2} \right) > 0 \implies \log_{2} \left( \frac{3x-1}{3x+2} \right) < 0 \]
This implies:
\[ \frac{3x-1}{3x+2} < 1 \]

### Step 7: Solve the inequality
Solving \( \frac{3x-1}{3x+2} < 1 \):
\[ 3x - 1 < 3x + 2 \implies -1 < 2 \quad \text{(always true)} \]

### Step 8: Combine all conditions
The domain of \( f(x) \) is determined by the intersection of the intervals:
\[ (-\infty, -\frac{2}{3}) \cup \left(\frac{1}{3}, \infty\right) \]

Thus, the final domain of the function \( f(x) \) is:
\[ \boxed{(-\infty, -\frac{2}{3}) \cup \left(\frac{1}{3}, \infty\right)} \]