Find the domain of definitions of the following function: `f(x)=log_(10)(1-log_(10)(x^(2)-5x+16))`

Explanation

To find the domain of the function \( f(x) = \log_{10}(1 - \log_{10}(x^2 - 5x + 16)) \), we need to ensure that the expression inside the logarithm is positive. Let's break down the steps:

### Step 1: Ensure the inner logarithm is defined
The logarithm \( \log_{10}(x^2 - 5x + 16) \) is defined when its argument is positive:
\[ x^2 - 5x + 16 > 0 \]

### Step 2: Analyze the quadratic expression
To analyze the quadratic \( x^2 - 5x + 16 \), we can calculate its discriminant:
\[ D = b^2 - 4ac = (-5)^2 - 4 \cdot 1 \cdot 16 = 25 - 64 = -39 \]
Since the discriminant is negative, the quadratic has no real roots and is always positive (as the coefficient of \( x^2 \) is positive).

### Step 3: Ensure the outer logarithm is defined
Next, we need to ensure that the argument of the outer logarithm is positive:
\[ 1 - \log_{10}(x^2 - 5x + 16) > 0 \]
This simplifies to:
\[ \log_{10}(x^2 - 5x + 16) < 1 \]

### Step 4: Convert logarithmic inequality to exponential form
Converting the logarithmic inequality to exponential form gives us:
\[ x^2 - 5x + 16 < 10 \]

### Step 5: Rearranging the inequality
Rearranging this inequality:
\[ x^2 - 5x + 6 < 0 \]

### Step 6: Factor the quadratic expression
Now we can factor the quadratic:
\[ x^2 - 5x + 6 = (x - 2)(x - 3) < 0 \]

### Step 7: Determine the intervals
To find the intervals where this product is negative, we can test the intervals defined by the roots \( x = 2 \) and \( x = 3 \):
- For \( x < 2 \), both factors are negative, hence the product is positive.
- For \( 2 < x < 3 \), the first factor is positive and the second is negative, hence the product is negative.
- For \( x > 3 \), both factors are positive, hence the product is positive.

### Step 8: Conclusion on the domain
Thus, the solution to the inequality \( (x - 2)(x - 3) < 0 \) gives us:
\[ 2 < x < 3 \]
Therefore, the domain of the function \( f(x) \) is:
\[ \text{Domain: } (2, 3) \]