Question

Solve the inequality: \( \log_2(x^2 - 5x + 6) >1 \)

• A. \( x \in (2,3) \cup (3, \infty) \)
• B. \( x \in (0,1) \cup (4, \infty)) \)
• C. \( x \in (0,2) \cup (2,3) \)
• D. \( x \in (1,2) \cup (3, \infty) \)

Hint:

\textbf{Tip:} Always consider both the inequality and the domain restrictions when solving logarithmic inequalities.

Answer 1

The Correct Option is B

To solve the inequality \(\log_2(x^2 - 5x + 6) > 1\), we need to follow these steps: 

1. Identify the domain for the logarithm: The expression inside the logarithm, \(x^2 - 5x + 6\), must be positive since the logarithm of a non-positive number is undefined. Thus, we solve the inequality:

\(x^2 - 5x + 6 > 0\).

Factoring the quadratic, we get:

\((x-2)(x-3) > 0\).

This inequality is satisfied when \(x \in (-\infty, 2) \cup (3, \infty)\).

2. Solve the logarithmic inequality:

\(\log_2(x^2 - 5x + 6) > 1\)

Convert the inequality from logarithmic form to exponential form:

\(x^2 - 5x + 6 > 2^1\)

\(x^2 - 5x + 6 > 2\)

Simplifying, we get:

\(x^2 - 5x + 4 > 0\).

Factor the quadratic:

\((x-1)(x-4) > 0\).

This inequality holds for \(x \in (-\infty, 1) \cup (4, \infty)\).

3. Combine the constraints: We need values of \(x\) that satisfy both \(x \in (-\infty, 2) \cup (3, \infty)\) and \(x \in (-\infty, 1) \cup (4, \infty)\). The intersection is:

\(x \in (0, 1) \cup (4, \infty)\).

Answer 2

Solving the Logarithmic Inequality: \( \log_2(x^2 - 5x + 6) > 1 \)

Step 1: Understanding the Given Inequality

We are given a logarithmic inequality: \[ \log_2(x^2 - 5x + 6) > 1 \] This inequality holds when the argument of the logarithm is greater than the base raised to the right-hand side: \[ x^2 - 5x + 6 > 2^1 = 2 \]

Step 2: Solve the Quadratic Inequality

Simplify and solve: \[ x^2 - 5x + 6 > 2 \Rightarrow x^2 - 5x + 4 > 0 \] Factor the quadratic: \[ (x - 4)(x - 1) > 0 \] This inequality holds for: \[ x < 1 \quad \text{or} \quad x > 4 \]

Step 3: Consider the Domain of the Logarithmic Function

For the logarithmic expression \( \log_2(x^2 - 5x + 6) \) to be defined, the argument must be positive: \[ x^2 - 5x + 6 > 0 \] Factor and solve: \[ (x - 2)(x - 3) > 0 \Rightarrow x < 2 \quad \text{or} \quad x > 3 \]

Step 4: Combine Both Conditions

From Step 2 (inequality): \( x < 1 \) or \( x > 4 \)
From Step 3 (domain): \( x < 2 \) or \( x > 3 \)

Combine these intervals using intersection (AND condition):

• For \( x < 1 \), it is already within the domain \( x < 2 \): valid.
• For \( x > 4 \), it also satisfies \( x > 3 \): valid

Final Answer

The solution set is: \[ x \in (-\infty, 1) \cup (4, \infty) \] Therefore, the final answer is: \( \boxed{x \in (-\infty, 1) \cup (4, \infty)} \)

Conclusion

Solving logarithmic inequalities requires two steps: solving the inequality itself and ensuring the domain conditions for the logarithm are satisfied. Always combine both sets of conditions for the final answer.