If `S={x:sqrt(log_xsqrt(3x))`, where `log_3xgt-1}`, then

Options

A. S is a finite set
B. S is an empty set
C. `Ssubset(0,infty)`
D. S properly contains `(1/3,infty)`

Answer

To solve the problem step by step, we need to analyze the given set \( S = \{ x : \sqrt{\log_x \sqrt{3x}} \} \) under the condition that \( \log_3 x > -1 \).

### Step 1: Rewrite the expression inside the square root
We start with the expression \( \sqrt{\log_x \sqrt{3x}} \). We can use the properties of logarithms to simplify this expression.

\[ \log_x \sqrt{3x} = \log_x (3x)^{1/2} = \frac{1}{2} \log_x (3x) = \frac{1}{2} (\log_x 3 + \log_x x) \]

Since \( \log_x x = 1 \), we can write:

\[ \log_x \sqrt{3x} = \frac{1}{2} (\log_x 3 + 1) \]

### Step 2: Substitute back into the original expression
Now substituting back into the square root:

\[ \sqrt{\log_x \sqrt{3x}} = \sqrt{\frac{1}{2} (\log_x 3 + 1)} \]

### Step 3: Analyze the condition \( \log_3 x > -1 \)
The condition \( \log_3 x > -1 \) can be rewritten using the change of base formula:

\[ \frac{\log x}{\log 3} > -1 \]

Multiplying both sides by \( \log 3 \) (which is positive), we get:

\[ \log x > -\log 3 \]

This implies:

\[ x > \frac{1}{3} \]

### Step 4: Rewrite the logarithm in terms of base \( x \)
Now, we can rewrite \( \log_x 3 \):

\[ \log_x 3 = \frac{\log 3}{\log x} \]

### Step 5: Substitute into the square root expression
Substituting this back into our expression for \( S \):

\[ \sqrt{\frac{1}{2} \left(\frac{\log 3}{\log x} + 1\right)} \]

### Step 6: Determine when the expression is defined
The expression \( \sqrt{\frac{1}{2} \left(\frac{\log 3}{\log x} + 1\right)} \) is defined when:

\[ \frac{1}{2} \left(\frac{\log 3}{\log x} + 1\right) \geq 0 \]

This implies:

\[ \frac{\log 3}{\log x} + 1 \geq 0 \]

### Step 7: Solve the inequality
Rearranging gives:

\[ \frac{\log 3}{\log x} \geq -1 \]

This leads to:

\[ \log 3 \geq -\log x \]

or equivalently:

\[ \log 3 + \log x \geq 0 \implies \log(3x) \geq 0 \implies 3x \geq 1 \implies x \geq \frac{1}{3} \]

### Conclusion
Combining the conditions \( x > \frac{1}{3} \) and \( x \geq \frac{1}{3} \), we find that \( S \) contains values in the interval \( \left(\frac{1}{3}, \infty\right) \).

Thus, we conclude that \( S \) properly contains \( \left(\frac{1}{3}, \infty\right) \).