The domain of definition of `f(x) = log {{log x)^2 - 5 log x+6}` is equal to

Options

A. `(0,10^2)`
B. `(10^3 ,oo)`
C. `(10^2 ,10^3)`
D. `(0,10^2 ) uu (10^3 ,oo)`

Answer

To find the domain of the function \( f(x) = \log((\log x)^2 - 5 \log x + 6) \), we need to ensure that the expression inside the logarithm is greater than zero and that \( x \) itself is positive.

### Step 1: Ensure \( x > 0 \)
Since we have \( \log x \) in the function, we need \( x \) to be greater than 0:
\[ x > 0 \]

### Step 2: Solve the inequality \( (\log x)^2 - 5 \log x + 6 > 0 \)
Let \( y = \log x \). Then, we need to solve the quadratic inequality:
\[ y^2 - 5y + 6 > 0 \]

### Step 3: Factor the quadratic
We can factor the quadratic:
\[ y^2 - 5y + 6 = (y - 2)(y - 3) \]
Thus, we need to solve:
\[ (y - 2)(y - 3) > 0 \]

### Step 4: Find the critical points
The critical points are \( y = 2 \) and \( y = 3 \). We will test intervals around these points to determine where the inequality holds.

### Step 5: Test intervals
1. **Interval \( (-\infty, 2) \)**: Choose \( y = 1 \):
\[ (1 - 2)(1 - 3) = (-1)(-2) = 2 > 0 \quad \text{(True)} \]
2. **Interval \( (2, 3) \)**: Choose \( y = 2.5 \):
\[ (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0 \quad \text{(False)} \]
3. **Interval \( (3, \infty) \)**: Choose \( y = 4 \):
\[ (4 - 2)(4 - 3) = (2)(1) = 2 > 0 \quad \text{(True)} \]

### Step 6: Combine intervals
The solution to the inequality \( (y - 2)(y - 3) > 0 \) is:
\[ y < 2 \quad \text{or} \quad y > 3 \]
Substituting back for \( y = \log x \):
1. \( \log x < 2 \) implies \( x < 10^2 = 100 \)
2. \( \log x > 3 \) implies \( x > 10^3 = 1000 \)

### Step 7: Combine with the condition \( x > 0 \)
Thus, the combined conditions for the domain are:
\[ x \in (0, 100) \cup (1000, \infty) \]

### Final Answer
The domain of definition of \( f(x) \) is:
\[ (0, 100) \cup (1000, \infty) \]