The value of `x, log_(1/2)x >= log_(1/3)x` is

Options

A. ` x in (0, 1)`
B. `x in [ 0,1)`
C. `x in (0,1]`
D. `x in [ 0, 1]`

Answer

To solve the inequality \( \log_{1/2} x \geq \log_{1/3} x \), we will follow these steps:

### Step 1: Rewrite the logarithmic inequality
We can rewrite the inequality using the change of base formula:
\[ \log_{1/2} x = \frac{\log x}{\log(1/2)} \quad \text{and} \quad \log_{1/3} x = \frac{\log x}{\log(1/3)} \]
Thus, the inequality becomes:
\[ \frac{\log x}{\log(1/2)} \geq \frac{\log x}{\log(1/3)} \]

### Step 2: Analyze the logarithmic bases
Since \( \log(1/2) < 0 \) and \( \log(1/3) < 0 \), we can multiply both sides of the inequality by \( \log(1/2) \) and \( \log(1/3) \) (keeping in mind that this reverses the inequality):
\[ \log x \cdot \log(1/3) \geq \log x \cdot \log(1/2) \]
This simplifies to:
\[ \log x \cdot \log(1/3) - \log x \cdot \log(1/2) \geq 0 \]
Factoring out \( \log x \):
\[ \log x \left( \log(1/3) - \log(1/2) \right) \geq 0 \]

### Step 3: Determine the sign of the factor
Next, we need to evaluate the sign of \( \log(1/3) - \log(1/2) \):
\[ \log(1/3) - \log(1/2) = \log\left(\frac{1/3}{1/2}\right) = \log\left(\frac{2}{3}\right) \]
Since \( \frac{2}{3} < 1 \), we have \( \log\left(\frac{2}{3}\right) < 0 \).

### Step 4: Solve the inequality
Now, we have:
\[ \log x \cdot \text{(a negative number)} \geq 0 \]
This implies:
\[ \log x \leq 0 \]
Thus, we find:
\[ x \leq 1 \]

### Step 5: Determine the domain of \( x \)
Since we are dealing with logarithms, \( x \) must be greater than 0:
\[ 0 < x \leq 1 \]

### Final Answer
The solution set for the inequality \( \log_{1/2} x \geq \log_{1/3} x \) is:
\[ x \in (0, 1] \]