The domain of the function: `f(x)=log_(3) [-(log_(3) x)^(2)+5 log_3x-6]" is :"`

Options

A. `(0,9) cup (27, oo)`
B. [9,27]
C. (9,27)
D. None of these

Answer

To find the domain of the function \( f(x) = \log_3 [-(\log_3 x)^2 + 5 \log_3 x - 6] \), we need to ensure that the argument of the logarithm is positive. Let's go through the steps systematically.

### Step 1: Set the argument of the logarithm greater than zero
The argument of the logarithm must be positive:
\[ -(\log_3 x)^2 + 5 \log_3 x - 6 > 0 \]

### Step 2: Rearrange the inequality
Rearranging the inequality gives:
\[ (\log_3 x)^2 - 5 \log_3 x + 6 < 0 \]

### Step 3: Let \( t = \log_3 x \)
Substituting \( t \) for \( \log_3 x \), we rewrite the inequality:
\[ t^2 - 5t + 6 < 0 \]

### Step 4: Factor the quadratic expression
We can factor the quadratic expression:
\[ (t - 2)(t - 3) < 0 \]

### Step 5: Determine the intervals using a number line
To find where the product is negative, we identify the critical points \( t = 2 \) and \( t = 3 \). We can test intervals around these points:
- For \( t < 2 \): Choose \( t = 1 \) → \( (1 - 2)(1 - 3) = (negative)(negative) = positive \)
- For \( 2 < t < 3 \): Choose \( t = 2.5 \) → \( (2.5 - 2)(2.5 - 3) = (positive)(negative) = negative \)
- For \( t > 3 \): Choose \( t = 4 \) → \( (4 - 2)(4 - 3) = (positive)(positive) = positive \)

Thus, the inequality \( (t - 2)(t - 3) < 0 \) holds true for:
\[ 2 < t < 3 \]

### Step 6: Convert back to \( x \)
Since \( t = \log_3 x \), we convert the inequality back:
\[ 2 < \log_3 x < 3 \]

### Step 7: Exponentiate to solve for \( x \)
Exponentiating gives:
\[ 3^2 < x < 3^3 \]
This simplifies to:
\[ 9 < x < 27 \]

### Step 8: Consider the domain condition
We also need to ensure \( x > 0 \) since logarithms are only defined for positive values. However, since \( 9 < x < 27 \) already implies \( x > 0 \), we do not need to add any further restrictions.

### Conclusion
The domain of the function \( f(x) \) is:
\[ (9, 27) \]