Problem 64 - Algebra 2

Question

PREREQUISITE SKILL Solve each equation or inequality. Check your solutions. $$ \log _{2} 3 x>\log _{2} 5 $$

Step-by-Step Solution

Answer

\( x > \frac{5}{3} \)

Step 1: Understand the Inequality

The given inequality is \( \log_{2} (3x) > \log_{2} (5) \). We need to determine the values of \( x \) for which this inequality holds true. Since the base of the logarithm (2) is greater than 1, the inequality \( \log_{b} (A) > \log_{b} (B) \) implies \( A > B \) when \( b > 1 \).

Step 2: Equate the Contents of the Logarithms

Based on the properties of logarithms, since \( \log_{2} (3x) > \log_{2} (5) \), it follows that \( 3x > 5 \).

Step 3: Solve the Inequality for x

To isolate \( x \), divide both sides of the inequality \( 3x > 5 \) by 3. This gives:\[ x > \frac{5}{3} \]

Step 4: Check the Solution

To verify this solution, choose a value of \( x \) that is greater than \( \frac{5}{3} \), for example \( x = 2 \). Check whether \( \log_{2} (3 \times 2) > \log_{2} (5) \):- \( \log_{2} (6) \approx 2.585 \)- \( \log_{2} (5) \approx 2.322 \)Since \( 2.585 > 2.322 \), our solution \( x > \frac{5}{3} \) is correct.

Key Concepts

Logarithmic Inequalities

When solving logarithmic inequalities, the goal is to determine the range of variable values for which the inequality is true. In our example, we have the inequality \( \log_{2} (3x) > \log_{2} (5) \). This type of inequality compares two logarithmic expressions with the same base, which in this case is 2.

Since the base of the logarithm is greater than 1, the inequality \( \log_{b} (A) > \log_{b} (B) \) implies \( A > B \). This is a fundamental property of logarithmic functions when the base is greater than 1, allowing us to drop the logarithms and directly compare the arguments inside.

• Ensure the base of the logarithm is greater than 1 to safely apply this property.
• Solve for the expression inside the logarithms after dropping the log.

In our example, \( 3x > 5 \) results after applying this property.

Properties of Logarithms

Logarithms have specific properties that become very useful in solving equations and inequalities. One important property is that logarithmic functions are increasing when the base is greater than 1. This property allows us to convert inequalities involving logarithms into inequalities involving their arguments.

Consider the following properties that are crucial:

• Equality Property: \( \log_{b} (A) = \log_{b} (B) \Rightarrow A = B \) provided \( A, B > 0 \) and \( b > 0 \), \( b eq 1 \).
• Inequality Property: \( \log_{b} (A) > \log_{b} (B) \Rightarrow A > B \) if \( b > 1 \).

Using these properties, we simplified \( \log_{2} (3x) > \log_{2} (5) \) to \( 3x > 5 \), and further found \( x > \frac{5}{3} \) by solving algebraically. These fundamental properties should always be checked for validity in context.

Verification of Solutions

Verifying solutions is a critical step in confirming the correctness of your result. For inequalities, this means checking if your solution set holds true when placed back into the original inequality.

To verify our solution \( x > \frac{5}{3} \), we selected a test value for \( x \), specifically \( x = 2 \), which is greater than \( \frac{5}{3} \). By substituting \( x = 2 \) back into the inequality \( \log_{2} (3 \times 2) > \log_{2} (5) \), we calculated:

• \( \log_{2} (6) \approx 2.585 \)
• \( \log_{2} (5) \approx 2.322 \)

The computed values show \( 2.585 > 2.322 \), confirming our initial solution is correct.

When verifying:

• Choose test values close to your boundary solutions to ensure accuracy.
• Double-check calculations to ensure that the inequality holds.

Verification not only confirms correctness but also builds confidence in handling similar problems.