Find the domain of each of the following functions: `f(x)=sqrt((log_(2)(x-2))/(log_(1//2)(3x-1)))`

Explanation

To find the domain of the function \( f(x) = \sqrt{\frac{\log_2(x-2)}{\log_{1/2}(3x-1)}} \), we need to ensure that the expression inside the square root is non-negative and that the logarithmic functions are defined.

### Step 1: Determine the conditions for the logarithms

1. **Logarithm Conditions**:
- For \( \log_2(x-2) \) to be defined, we need:
\[ x - 2 > 0 \implies x > 2 \]
- For \( \log_{1/2}(3x-1) \) to be defined, we need:
\[ 3x - 1 > 0 \implies x > \frac{1}{3} \]

### Step 2: Analyze the square root condition

2. **Square Root Condition**:
- The expression inside the square root must be non-negative:
\[ \frac{\log_2(x-2)}{\log_{1/2}(3x-1)} \geq 0 \]
- This fraction is non-negative when both the numerator and denominator are either both positive or both negative.

### Step 3: Solve the inequalities

3. **Case 1: Both positive**:
- \( \log_2(x-2) \geq 0 \implies x - 2 \geq 1 \implies x \geq 3 \)
- \( \log_{1/2}(3x-1) > 0 \implies 3x - 1 < 1 \implies 3x < 2 \implies x < \frac{2}{3} \)
- This case gives no valid solution since \( x \geq 3 \) and \( x < \frac{2}{3} \) cannot both be true.

4. **Case 2: Both negative**:
- \( \log_2(x-2) < 0 \implies x - 2 < 1 \implies x < 3 \)
- \( \log_{1/2}(3x-1) < 0 \implies 3x - 1 > 1 \implies 3x > 2 \implies x > \frac{2}{3} \)
- This case gives us \( \frac{2}{3} < x < 3 \).

### Step 4: Combine conditions

5. **Combine conditions**:
- From the logarithmic conditions, we have \( x > 2 \) and from the square root conditions, we have \( \frac{2}{3} < x < 3 \).
- The intersection of these conditions is:
\[ 2 < x < 3 \]

### Conclusion

Thus, the domain of the function \( f(x) \) is:
\[ \text{Domain} = (2, 3) \]