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Applications of Logarithms

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Apply logarithms to solve exponential equations and inequalities.

Applications of Logarithms

Logarithms are the inverses of exponential functions and have wide applications in solving equations, inequalities, and real-world problems involving growth, decay, and scales. This section covers the key concepts, properties, and applications of logarithms, along with essential set theory basics that often appear in entrance exams.

1. Logarithmic Equations

A logarithmic equation is an equation that involves logarithms of expressions. The general form is:

\[ \log_b x = y \quad \Longleftrightarrow \quad b^y = x \]

where \( b \) is the base (positive and \( b \neq 1 \)), \( x > 0 \), and \( y \) is any real number.

Key Properties of Logarithms

  • Product Rule: \( \log_b (MN) = \log_b M + \log_b N \)
  • Quotient Rule: \( \log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N \)
  • Power Rule: \( \log_b (M^k) = k \log_b M \)
  • Change of Base Formula: \( \log_b a = \frac{\log_c a}{\log_c b} \), where \( c \) is any positive base \( \neq 1 \)

These properties allow us to simplify and solve logarithmic equations effectively.

Solving Logarithmic Equations

To solve equations involving logarithms:

  1. Ensure the arguments of all logarithms are positive (domain restrictions).
  2. Use logarithmic properties to combine or expand logarithms.
  3. Convert the logarithmic equation to exponential form if needed.
  4. Solve the resulting algebraic equation.
  5. Check for extraneous solutions that violate domain restrictions.

2. Logarithmic Inequalities

Logarithmic inequalities involve expressions like \( \log_b f(x) > k \) or \( \log_b f(x) \leq k \). Solving these requires careful attention to the domain and the base of the logarithm.

Rules for Solving Logarithmic Inequalities

  • If \( b > 1 \), then \( \log_b f(x) > k \iff f(x) > b^k \)
  • If \( 0 < b < 1 \), then \( \log_b f(x) > k \iff f(x) < b^k \)
  • Always ensure \( f(x) > 0 \) because the argument of a logarithm must be positive.

Graphical interpretation helps visualize the solution set by plotting \( y = \log_b f(x) \) and \( y = k \).

3. Applications of Logarithms

Logarithms are used in many real-world contexts:

  • Exponential Growth and Decay: Population growth, radioactive decay, and compound interest problems often use logarithms to solve for time or rate.
  • pH Scale: pH is defined as \( \text{pH} = -\log_{10} [H^+] \), measuring acidity.
  • Richter Scale: Measures earthquake magnitude logarithmically.
  • Compound Interest: Logarithms help find the time needed for an investment to grow to a certain amount.

4. Set Theory Basics Relevant to Logarithms

Understanding sets and their properties is crucial for entrance exams, especially when dealing with problems involving cardinality, subsets, and power sets.

Key Concepts

  • Subset: \( A \subseteq B \) means every element of \( A \) is in \( B \).
  • Null Set (Empty Set): Denoted \( \emptyset \), it is a subset of every set.
  • Power Set: The set of all subsets of a set \( S \), denoted \( \mathcal{P}(S) \), has \( 2^{n} \) elements if \( n = |S| \).
  • Set Operations: Union \( A \cup B \), intersection \( A \cap B \), difference \( A - B \), and symmetric difference \( A \oplus B \).
  • Cardinality: Number of elements in a set, denoted \( n(A) \).
  • Equivalent Sets: Sets with the same cardinality.

These concepts often appear in problems that combine algebraic and set theory reasoning.

x y y = k x = b^k

Graph of \( y = \log_b x \) and line \( y = k \). For \( b > 1 \), \( \log_b x > k \) when \( x > b^k \).

Exam Tip: Always check the domain of logarithmic expressions before solving. Ignoring domain restrictions is a common mistake.

Worked Examples

Example 1 (Easy)

Problem: Solve \( \log_2 (x - 1) = 3 \).

Solution:

  1. Rewrite in exponential form: \( x - 1 = 2^3 = 8 \).
  2. So, \( x = 8 + 1 = 9 \).
  3. Check domain: \( x - 1 > 0 \Rightarrow x > 1 \). Since \( 9 > 1 \), solution is valid.

Answer: \( x = 9 \).

Example 2 (Easy)

Problem: Solve \( \log_3 (x^2 - 4) = 2 \).

Solution:

  1. Rewrite: \( x^2 - 4 = 3^2 = 9 \).
  2. So, \( x^2 = 13 \Rightarrow x = \pm \sqrt{13} \).
  3. Check domain: \( x^2 - 4 > 0 \Rightarrow x^2 > 4 \Rightarrow |x| > 2 \).
  4. Both \( \sqrt{13} \approx 3.6 \) and \( -\sqrt{13} \approx -3.6 \) satisfy domain.

Answer: \( x = \pm \sqrt{13} \).

Example 3 (Medium)

Problem: Solve \( \log_5 (x + 4) + \log_5 (x - 1) = 1 \).

Solution:

  1. Use product rule: \( \log_5 [(x + 4)(x - 1)] = 1 \).
  2. Rewrite: \( (x + 4)(x - 1) = 5^1 = 5 \).
  3. Expand: \( x^2 - x + 4x - 4 = 5 \Rightarrow x^2 + 3x - 4 = 5 \).
  4. Bring all terms to one side: \( x^2 + 3x - 9 = 0 \).
  5. Solve quadratic: \( x = \frac{-3 \pm \sqrt{9 + 36}}{2} = \frac{-3 \pm \sqrt{45}}{2} = \frac{-3 \pm 3\sqrt{5}}{2} \).
  6. Check domain: \( x + 4 > 0 \Rightarrow x > -4 \), and \( x - 1 > 0 \Rightarrow x > 1 \). So, \( x > 1 \).
  7. Evaluate roots:
    • \( \frac{-3 + 3\sqrt{5}}{2} \approx \frac{-3 + 6.7}{2} = \frac{3.7}{2} = 1.85 > 1 \) valid.
    • \( \frac{-3 - 3\sqrt{5}}{2} \approx \frac{-3 - 6.7}{2} = \frac{-9.7}{2} = -4.85 < 1 \) invalid.

Answer: \( x = \frac{-3 + 3\sqrt{5}}{2} \).

Example 4 (Medium)

Problem: Solve the inequality \( \log_2 (x - 2) \leq 3 \).

Solution:

  1. Domain: \( x - 2 > 0 \Rightarrow x > 2 \).
  2. Rewrite inequality: \( \log_2 (x - 2) \leq 3 \Rightarrow x - 2 \leq 2^3 = 8 \).
  3. So, \( x \leq 10 \).
  4. Combine domain and inequality: \( 2 < x \leq 10 \).

Answer: \( x \in (2, 10] \).

Example 5 (Hard)

Problem: If \( n(A) = 37 \), \( n(B) = 25 \), and \( n(A \cup B) = 50 \), find \( n(A \cap B) \).

Solution:

  1. Recall the formula: \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \).
  2. Substitute values: \( 50 = 37 + 25 - n(A \cap B) \).
  3. Simplify: \( 50 = 62 - n(A \cap B) \Rightarrow n(A \cap B) = 62 - 50 = 12 \).

Answer: \( n(A \cap B) = 12 \).

Example 6 (Hard)

Problem: Find the maximum number of elements in \( A \oplus B \) (symmetric difference) if \( n(A) = 5 \) and \( n(B) = 4 \).

Solution:

  1. Recall symmetric difference: \( A \oplus B = (A - B) \cup (B - A) \).
  2. Maximum symmetric difference occurs when \( A \cap B = \emptyset \) (no common elements).
  3. Then, \( n(A \oplus B) = n(A) + n(B) = 5 + 4 = 9 \).

Answer: Maximum \( n(A \oplus B) = 9 \).

Formula Bank

  • Logarithm Definition: \( \log_b x = y \iff b^y = x \), \( b > 0, b \neq 1, x > 0 \)
  • Product Rule: \( \log_b (MN) = \log_b M + \log_b N \)
  • Quotient Rule: \( \log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N \)
  • Power Rule: \( \log_b (M^k) = k \log_b M \)
  • Change of Base Formula: \( \log_b a = \frac{\log_c a}{\log_c b} \)
  • Set Union and Intersection: \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
  • Power Set Cardinality: If \( |S| = n \), then \( |\mathcal{P}(S)| = 2^n \)
  • Symmetric Difference: \( A \oplus B = (A - B) \cup (B - A) \)
  • Domain of Logarithm: Argument \( > 0 \)
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