Problem 37 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\log _{9}\left(x^{2}-5 x+6\right)>\log _{3}(x-4), x\) belongs to (A) \((-\infty, 4)\) (B) \((4, \infty)\) (C) \((-\infty,-4) \cup(4, \infty)\) (D) no real value of \(x\)

Step-by-Step Solution

Answer

The answer is (B) \((4, \infty)\).

Step 1: Express both sides with the same base

To compare the two logarithmic expressions, let's express them using the same base. Recall that \( 9 = 3^2 \), so we apply the change of base formula: \( \log_{9}(x^2 - 5x + 6) = \log_{3}(x^2 - 5x + 6) / 2 \). Now the inequality becomes:\[ \frac{\log_{3}(x^2 - 5x + 6)}{2} > \log_{3}(x - 4) \]Multiply both sides by 2 for simplicity:\[ \log_{3}(x^2 - 5x + 6) > 2 \log_{3}(x - 4) \]

Step 2: Simplify and solve the inequality

Use the property of logarithms \( a\log_{b}c = \log_{b}(c^a) \) on the right-hand side:\[ \log_{3}(x^2 - 5x + 6) > \log_{3}((x - 4)^2) \]Now, remove the logs by setting the arguments in the inequality since logs are increasing:\[ x^2 - 5x + 6 > (x - 4)^2 \]

Step 3: Expand and combine terms

Expand the right-hand side:\[ x^2 - 5x + 6 > x^2 - 8x + 16 \]Cancel \(x^2\) from both sides:\[ -5x + 6 > -8x + 16 \]

Step 4: Solve for x

Rearrange the inequality to isolate \(x\):\[ 3x > 10 \]Divide both sides by 3:\[ x > \frac{10}{3} \]

Step 5: Consider the domain of the logarithmic functions

Ensure the domain of each log is valid: \(x^2 - 5x + 6 > 0\) and \(x - 4 > 0\). The first expression factorizes as \((x - 2)(x - 3) > 0\), which is true for \(x < 2\) or \(x > 3\). From \(x - 4 > 0\), we have \(x > 4\). Thus, the valid solutions are in \(x > 4\).

Step 6: Determine the final answer

Combining these insights, the final answer is based on the domain and the solution constraints from the inequality: \(x > 4\). Therefore, the answer is (B) \((4, \infty)\).

Key Concepts

Logarithmic Inequalities

Logarithmic inequalities involve comparing two logarithmic expressions. For the inequality \( \log_{9}\left(x^{2}-5x+6\right)>\log_{3}(x-4) \), we start by making the bases of the logarithms the same. This ensures a straightforward comparison. When dealing with inequalities, remember that if the base is greater than 1, the logarithmic function is increasing. Thus, greater arguments lead to greater logarithmic values.

Transform the inequality so both sides have the same base using the change of base formula. This step involves a critical principle: if \( \log_{b}(A) > \log_{b}(B) \), then \( A > B \) provided \(A, B > 0\). Once this transformation is complete, solve the simpler inequality in the arguments. Simplify it to isolate \(x\).

• Express both logs with a common base.
• Use the properties of logarithms to simplify.
• Set the arguments of the logs greater for comparison and solution.

This process highlights the fundamental properties of logarithms and requires close attention to the conditions that arise from their domains.

Change of Base Formula

In logarithmic functions, the change of base formula allows you to convert a logarithm with one base into an equivalent logarithm with another base. The formula is:
\[\log_{b}(x) = \frac{\log_{k}(x)}{\log_{k}(b)}\]which essentially divides the logarithm of the number by the logarithm of the base, both in the new base \(k\).

In many exercises, especially involving inequalities, expressing both sides of the comparison with the same base simplifies solving the problem. In our problem, converting \( \log_{9}(x^2-5x+6) \) into a base 3 logarithm is crucial. We use:

• \(9 = 3^2 \), hence, \( \log_{9}(\text{expression}) = \frac{\log_{3}(\text{expression})}{2} \).
• This transformation helps align the inequality into a comparable format.
• Once expressed in the same base, inequalities become straightforward to handle.

Using the change of base formula not only facilitates comparison but often reveals important simplifications about the logarithmic expressions involved.

Domain of Logarithmic Functions

Understanding the domain is critical when dealing with logarithms. The domain of a logarithmic function is the set of values for which the function is defined, meaning what \(x\) can be for \( \log_{b}(x) \) to be valid. Generally, \(x\) must be positive because the logarithm of a non-positive number is undefined.

Before solving any inequality involving logarithms, always check the domain conditions:

• The arguments of all logarithms must be positive.
• In our case: \(x^2 - 5x + 6 > 0\) and \(x - 4 > 0\).
• Factor and solve these inequalities to find valid \(x\).

Analyzing \(x^2 - 5x + 6 > 0\), we find the intervals where it holds by solving \((x-2)(x-3) > 0 \), leading to solutions \(x < 2\) and \(x > 3\).
For \(x - 4 > 0\), it simplifies to \(x > 4\). The final domain constraint comes from taking the intersection of these results, ensuring that \(x\) adheres to all conditions: \(x > 4\). This domain consideration complements the final interval solution for \(x\).