If `log_(3)x-(log_(3)x)^(2)le(3)/(2)log_((1//2sqrt(2)))4`, then x can belong to (a) `(-oo,1//3)` (b) `(9,oo)` (c) (1,6) (d) `(-oo,0)`

Options

A. `(-oo,1//3)`
B. `(9,oo)`
C. (1,6)
D. `(-oo,0)`

Explanation

To solve the inequality \( \log_3 x - (\log_3 x)^2 \leq \frac{3}{2} \log_{\frac{1}{2\sqrt{2}}} 4 \), we will follow these steps:

### Step 1: Simplify the Right Side
First, we simplify the right-hand side of the inequality.

\[ \log_{\frac{1}{2\sqrt{2}}} 4 = \frac{\log 4}{\log \left(\frac{1}{2\sqrt{2}}\right)} \]

Using the properties of logarithms, we can rewrite \( \log 4 \) as:

\[ \log 4 = \log(2^2) = 2 \log 2 \]

Now, we simplify \( \log \left(\frac{1}{2\sqrt{2}}\right) \):

\[ \log \left(\frac{1}{2\sqrt{2}}\right) = \log(1) - \log(2\sqrt{2}) = 0 - \left(\log 2 + \log \sqrt{2}\right) = -\log 2 - \frac{1}{2} \log 2 = -\frac{3}{2} \log 2 \]

Thus, we have:

\[ \log_{\frac{1}{2\sqrt{2}}} 4 = \frac{2 \log 2}{-\frac{3}{2} \log 2} = -\frac{4}{3} \]

Now substituting this back into the inequality gives:

\[ \log_3 x - (\log_3 x)^2 \leq -\frac{4}{3} \]

### Step 2: Rearranging the Inequality
Rearranging the inequality gives:

\[ (\log_3 x)^2 - \log_3 x - \frac{4}{3} \geq 0 \]

### Step 3: Let \( t = \log_3 x \)
Let \( t = \log_3 x \). The inequality now becomes:

\[ t^2 - t - \frac{4}{3} \geq 0 \]

### Step 4: Factor the Quadratic
To factor the quadratic, we can use the quadratic formula:

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + \frac{16}{3}}}{2} = \frac{1 \pm \sqrt{\frac{19}{3}}}{2} \]

Calculating the roots:

\[ t_1 = \frac{1 + \sqrt{\frac{19}{3}}}{2}, \quad t_2 = \frac{1 - \sqrt{\frac{19}{3}}}{2} \]

### Step 5: Determine the Intervals
The roots divide the number line into intervals. We need to test the sign of the quadratic in each interval:

1. \( (-\infty, t_2) \)
2. \( (t_2, t_1) \)
3. \( (t_1, \infty) \)

The quadratic opens upwards (since the coefficient of \( t^2 \) is positive), so it will be positive outside the roots and negative between them.

### Step 6: Find the Valid Intervals
Thus, the solution to the inequality is:

\[ t \leq t_2 \quad \text{or} \quad t \geq t_1 \]

### Step 7: Convert Back to \( x \)
Since \( t = \log_3 x \):

1. For \( t \leq t_2 \):
\[ x \leq 3^{t_2} \]
2. For \( t \geq t_1 \):
\[ x \geq 3^{t_1} \]

### Step 8: Analyze the Results
We need to find the numerical values of \( 3^{t_1} \) and \( 3^{t_2} \) to determine the ranges for \( x \).

### Final Answer
After evaluating the intervals, we find that \( x \) can belong to \( (9, \infty) \).