The greatest value of the function `log_(x) 1//9 - log_(3) x^(2) (x gt 1)` is

Options

A. 2
B. 0
C. `-4`
D. `-2`

Answer

To find the greatest value of the function \( f(x) = \log_{x} \left( \frac{1}{9} \right) - \log_{3} (x^2) \) for \( x > 1 \), we can follow these steps:

### Step 1: Rewrite the logarithmic expressions
Using the change of base formula for logarithms, we can rewrite the function:
\[ f(x) = \frac{\log \left( \frac{1}{9} \right)}{\log x} - 2 \log_{3} x \]
Since \( \log_{3} x = \frac{\log x}{\log 3} \), we can rewrite the second term as:
\[ -2 \log_{3} x = -2 \cdot \frac{\log x}{\log 3} = -\frac{2 \log x}{\log 3} \]
Thus, the function becomes:
\[ f(x) = \frac{\log \left( \frac{1}{9} \right)}{\log x} - \frac{2 \log x}{\log 3} \]

### Step 2: Simplify the function
We know that \( \log \left( \frac{1}{9} \right) = \log (9^{-1}) = -\log 9 = -2 \log 3 \). Therefore, we can substitute this into our function:
\[ f(x) = \frac{-2 \log 3}{\log x} - \frac{2 \log x}{\log 3} \]

### Step 3: Set \( y = \log x \)
Let \( y = \log x \). Then, \( x = 10^y \) and since \( x > 1 \), we have \( y > 0 \). The function now becomes:
\[ f(y) = \frac{-2 \log 3}{y} - \frac{2y}{\log 3} \]

### Step 4: Differentiate the function
To find the maximum value, we differentiate \( f(y) \):
\[ f'(y) = \frac{2 \log 3}{y^2} - \frac{2}{\log 3} \]
Setting \( f'(y) = 0 \):
\[ \frac{2 \log 3}{y^2} - \frac{2}{\log 3} = 0 \]
This simplifies to:
\[ \frac{2 \log 3}{y^2} = \frac{2}{\log 3} \]
Cross-multiplying gives:
\[ (\log 3)^2 = y^2 \implies y = \log 3 \quad (\text{since } y > 0) \]

### Step 5: Find \( x \)
Since \( y = \log x \), we have:
\[ \log x = \log 3 \implies x = 3 \]

### Step 6: Calculate the maximum value of \( f(x) \)
Now, substitute \( x = 3 \) back into the function:
\[ f(3) = \log_{3} \left( \frac{1}{9} \right) - \log_{3} (3^2) \]
Calculating each term:
\[ \log_{3} \left( \frac{1}{9} \right) = \log_{3} (3^{-2}) = -2 \]
\[ \log_{3} (3^2) = 2 \]
Thus,
\[ f(3) = -2 - 2 = -4 \]

### Final Answer
The greatest value of the function \( f(x) \) for \( x > 1 \) is:
\[ \boxed{-4} \]