The domain of definition of the function `f(x)=log_(2)[-(log_(2)x)^(2)+5log_(2)x-6]` , is

Options

A. `(4, 8)`
B. `[9,27]`
C. `(9,27)`
D. None of these

Answer

To find the domain of the function \( f(x) = \log_2[-(\log_2 x)^2 + 5 \log_2 x - 6] \), we need to ensure that the argument of the logarithm is positive. Let's go through the solution step by step:

### Step 1: Set the argument of the logarithm greater than zero
We start with the condition that the argument of the logarithm must be greater than zero:
\[ -(\log_2 x)^2 + 5 \log_2 x - 6 > 0 \]

### Step 2: Substitute \( t = \log_2 x \)
Let \( t = \log_2 x \). Then, we can rewrite the inequality as:
\[ -t^2 + 5t - 6 > 0 \]

### Step 3: Rearrange the inequality
Multiplying through by -1 (and reversing the inequality sign) gives:
\[ t^2 - 5t + 6 < 0 \]

### Step 4: Factor the quadratic expression
Next, we factor the quadratic:
\[ t^2 - 5t + 6 = (t - 2)(t - 3) \]
Thus, the inequality becomes:
\[ (t - 2)(t - 3) < 0 \]

### Step 5: Analyze the sign of the product
To find the intervals where this product is negative, we identify the roots \( t = 2 \) and \( t = 3 \). We can test the intervals:
- For \( t < 2 \): both factors are negative, so the product is positive.
- For \( 2 < t < 3 \): one factor is positive and the other is negative, so the product is negative.
- For \( t > 3 \): both factors are positive, so the product is positive.

Thus, the solution to the inequality is:
\[ 2 < t < 3 \]

### Step 6: Convert back to \( x \)
Now, we convert back to \( x \) using \( t = \log_2 x \):
- From \( t > 2 \): \( \log_2 x > 2 \) implies \( x > 2^2 = 4 \).
- From \( t < 3 \): \( \log_2 x < 3 \) implies \( x < 2^3 = 8 \).

### Step 7: Combine the results
Combining these inequalities gives us:
\[ 4 < x < 8 \]

### Conclusion
The domain of the function \( f(x) \) is:
\[ (4, 8) \]