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. This means we need to solve the inequality:

\[ -(\log_3 x)^2 + 5 \log_3 x - 6 > 0 \]

Let's define \( y = \log_3 x \). Then, we can rewrite the inequality as:

\[ -y^2 + 5y - 6 > 0 \]

Now, we can rearrange this to:

\[ y^2 - 5y + 6 < 0 \]

Next, we will factor the quadratic expression:

\[ y^2 - 5y + 6 = (y - 2)(y - 3) \]

Now, we need to find the intervals where this product is less than zero:

\[ (y - 2)(y - 3) < 0 \]

To find the critical points, we set each factor to zero:

1. \( y - 2 = 0 \) gives \( y = 2 \)
2. \( y - 3 = 0 \) gives \( y = 3 \)

Now we can test the intervals determined by these points: \( (-\infty, 2) \), \( (2, 3) \), and \( (3, \infty) \).

1. **For the interval \( (-\infty, 2) \)**:
- Choose \( y = 1 \): \( (1 - 2)(1 - 3) = (-1)(-2) = 2 > 0 \)

2. **For the interval \( (2, 3) \)**:
- Choose \( y = 2.5 \): \( (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0 \)

3. **For the interval \( (3, \infty) \)**:
- Choose \( y = 4 \): \( (4 - 2)(4 - 3) = (2)(1) = 2 > 0 \)

From this testing, we find that the inequality \( (y - 2)(y - 3) < 0 \) holds true in the interval \( (2, 3) \).

Thus, we have:

\[ 2 < y < 3 \]

Recalling that \( y = \log_3 x \), we can convert this back to \( x \):

1. For \( y = 2 \): \( \log_3 x = 2 \) implies \( x = 3^2 = 9 \)
2. For \( y = 3 \): \( \log_3 x = 3 \) implies \( x = 3^3 = 27 \)

Therefore, the domain of \( f(x) \) in terms of \( x \) is:

\[ 9 < x < 27 \]

**Final Answer: The domain of the function is \( (9, 27) \).**

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