Find the domain of the function `f(x)=log_10((log_10 x^2)-5log_10 x+6)`

Options

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

Answer

To find the domain of the function \( f(x) = \log_{10}(\log_{10}(x^2) - 5\log_{10}(x) + 6) \), we need to ensure that the expression inside the logarithm is greater than zero. Let's break down the solution step by step.

### Step 1: Define the inner function
Let \( t = \log_{10}(x) \). Then, we can rewrite the function as:
\[ f(x) = \log_{10}(t^2 - 5t + 6) \]

### Step 2: Set the condition for the logarithm
For \( f(x) \) to be defined, the argument of the logarithm must be greater than zero:
\[ t^2 - 5t + 6 > 0 \]

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

### Step 4: Determine the intervals
To find where the product is positive, we analyze the sign of the factors:
- The roots are \( t = 2 \) and \( t = 3 \).
- We test the intervals: \( (-\infty, 2) \), \( (2, 3) \), and \( (3, \infty) \).

1. For \( t < 2 \) (e.g., \( t = 1 \)):
\((1 - 2)(1 - 3) = (-1)(-2) = 2 > 0\) (positive)

2. For \( 2 < t < 3 \) (e.g., \( t = 2.5 \)):
\((2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0\) (negative)

3. For \( t > 3 \) (e.g., \( t = 4 \)):
\((4 - 2)(4 - 3) = (2)(1) = 2 > 0\) (positive)

Thus, the solution to \( (t - 2)(t - 3) > 0 \) is:
\[ t < 2 \quad \text{or} \quad t > 3 \]

### Step 5: Convert back to \( x \)
Recall that \( t = \log_{10}(x) \). We convert the intervals back to \( x \):
1. For \( t < 2 \):
\[ \log_{10}(x) < 2 \implies x < 10^2 = 100 \]

2. For \( t > 3 \):
\[ \log_{10}(x) > 3 \implies x > 10^3 = 1000 \]

### Step 6: Combine the intervals
The domain of \( f(x) \) is:
\[ (-\infty, 100) \cup (1000, \infty) \]

### Final Answer
Thus, the domain of the function \( f(x) \) is:
\[ (-\infty, 100) \cup (1000, \infty) \]