Problem 301 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \log _{2} \frac{x-3}{x+2}<0 $$

Step-by-Step Solution

Answer

The solution set for the given inequality \(\log _{2} \frac{x-3}{x+2}<0\) is \(x \in \left(-\infty, -2\right) \cup \left(-2, \infty\right)\). However, x cannot be -2 since that would make the denominator 0. Therefore, the solution set can also be written as \(x \in \mathbb{R} - \{-2\}\).

Step 1: Analyze and rewrite inequality

We are given \(\log _{2} \frac{x-3}{x+2}<0 \). To better analyze it, let's rewrite it as \( \log _{2} (x-3) - \log _{2} (x+2)<0 \).

Step 2: Exponentiate both sides

To remove the logarithms, we apply the base of 2 to both sides of the inequality: \[ 2^{\log _{2}(x-3)} = x-3 \\ 2^{\log _{2}(x+2)} = x+2 \\ \] Hence, the inequality becomes: \(\frac{x-3}{x+2} < 1\)

Step 3: Manipulate the inequality

Now, we want to get everything on one side of the inequality to analyze the critical points. We manipulate the inequality: \[\frac{x-3}{x+2} -1 < 0\] Combining the fractions, we get \(\frac{x-3 -(x+2)}{(x+2)}< 0\), which simplifies to the inequality: \(\frac{-5}{x+2}< 0\)

Step 4: Determine critical points and intervals

Next, we find all critical points and where the inequality is equal to 0. The critical point for this inequality is at \(x = -2\) (the denominator). Now, we create two intervals based on this critical point: Intervals: \(\left(-\infty, -2\right)\) and \(\left(-2, \infty\right)\)

Step 5: Test the intervals

Let's test the intervals by plugging in an x-value from each interval into the inequality \(\frac{-5}{x+2}< 0\), and see if the inequality holds. Interval 1: \(x \in \left(-\infty, -2\right)\) Choose \(x = -3\), plug in: \(\frac{-5}{-3+2} > 0\) (True) Interval 2: \(x \in \left(-2, \infty\right)\) Choose \(x = 0\), plug in: \(\frac{-5}{0+2} < 0\) (True) Hence, both intervals satisfy the inequality.

Step 6: Provide the solution set

Now, considering both intervals are true, the solution set for the given inequality is given by: \(x \in \left(-\infty, -2\right) \cup \left(-2, \infty\right)\) However, x cannot be -2 since that would make the denominator 0. Therefore, the solution set can also be written as: \(x \in \mathbb{R} - \{-2\}\)

Key Concepts

Exponentiation

Logarithmic inequalities often require us to perform exponentiation in order to solve them. Exponentiation is the process of raising a number to the power of another number. It is the inverse operation of logarithms. For instance, in the base 2 logarithm, the exponentiation would involve 2 as the base.
When dealing with inequalities that include logarithms, such as \(\log_2\frac{x-3}{x+2}<0\), exponentiating both sides helps us to remove the logarithm and simplify the expression. This is possible due to the fact that if \(b\) is a positive real number and \(b^y = x\), then \(y\) is the logarithm of \(x\) (with base \(b\)). In the step by step solution provided, exponentiating with base 2 transforms the logarithmic inequality into a rational inequality, which is easier to manipulate and solve.
Understanding exponentiation is crucial because it allows us to isolate the variable of interest in an inequality, making it possible to better understand and solve such mathematical problems.

Critical Points in Inequalities

In the context of solving inequalities, a 'critical point' is a value that could potentially change the inequality's truth value. Identifying these points is crucial for determining the solution set.
For the given inequality \(\frac{-5}{x+2}< 0\), the critical point is the value of \(x\) that equates the denominator to zero, in this case, \(x = -2\). This is because dividing by zero would yield an undefined expression, and hence, we cannot include this value in the solution set.

How to Find Critical Points

• Set the denominator and the numerator (if it contains variables) of the rational expression equal to zero.
• Solve each equation to find the potential critical points.

After determining the critical points, we divide the number line into intervals separated by these critical points and test the intervals. This method ensures that we examine the behavior of the inequality at values near those critical points, leading us to find where the inequality holds true.

Solution Sets for Inequalities

Once we have identified the critical points and tested the intervals, we can define the solution set for the inequality. The solution set is the range of values that satisfies the inequality.
In our exercise, after testing intervals created by the critical point (\(x = -2\)), we find that the two intervals \((-\infty, -2)\) and \((-2, \infty)\) satisfy the inequality. As such, the solution set includes all real numbers except the critical point where the function is undefined. This is denoted as \(x \in \mathbb{R} - \{-2\}\).

Key Points in Defining Solution Sets

The solution set should:

• Include all values that make the inequality true.
• Exclude any values that could make any part of the original equation undefined (like division by zero).
• Be expressed in the simplest form, often using interval notation or set-builder notation.

For complete understanding, students should practice converting the final solution into different forms and interpreting what each form means in the context of the inequality. Recognizing how the solution set is derived and how it's related to the initial inequality is essential for mastering logarithmic inequalities.