Logarithms
Introduction
Logarithms are the inverse operation of exponentiation. They are essential tools in science, engineering, and advanced mathematics. For ICSE Class 9, understanding the definition and laws of logarithms is fundamental.
Definition
If a^x = n, then logₐ n = x (where a > 0, a ≠ 1, and n > 0).
Read as: 'log of n to the base a equals x'
Examples:
- 2³ = 8 => log₂ 8 = 3
- 10² = 100 => log₁₀ 100 = 2
- 5⁻¹ = 1/5 => log₅ (1/5) = -1
- 10⁰ = 1 => log₁₀ 1 = 0
Key Points
- Base of a logarithm is always positive and not equal to 1
- The argument (n) must be positive
- logₐ 1 = 0 for any valid base a
- logₐ a = 1 for any valid base a
Laws of Logarithms
Law 1: Product Rule
logₐ (m × n) = logₐ m + logₐ n
<ICSEExample title="Express log(10 × 100) using product rule"> <Solution> log₁₀(10 × 100) = log₁₀ 10 + log₁₀ 100 = 1 + 2 = 3 Check: log₁₀ 1000 = 3 ✓ </Solution> </ICSEExample>Law 2: Quotient Rule
logₐ (m/n) = logₐ m — logₐ n
<ICSEExample title="Simplify log₂(32/4)"> <Solution> log₂(32/4) = log₂ 32 — log₂ 4 = 5 — 2 = 3 Check: log₂ 8 = 3 ✓ </Solution> </ICSEExample>Law 3: Power Rule
logₐ (m^n) = n × logₐ m
<ICSEExample title="Simplify log₃ 81⁴"> <Solution> log₃ 81⁴ = 4 × log₃ 81 = 4 × log₃ 3⁴ = 4 × 4 = 16 </Solution> </ICSEExample>Common Logarithms (Base 10)
Logarithms with base 10 are called common logarithms. They are widely used in calculations.
Notation: log₁₀ x is often written simply as log x.
Characteristic and Mantissa
For a positive number:
- Characteristic: the integer part (before decimal)
- Mantissa: the fractional part (after decimal, always positive)
Example: If log 235 = 2.3711:
- Characteristic = 2
- Mantissa = 0.3711
Antilogarithms
Antilogarithm is the inverse operation of logarithm. If log x = y, then antilog y = x.
<ICSEExample title="Find x if log x = 3.8451"> <Solution> x = antilog(3.8451) Characteristic = 3, so x has 4 digits before decimal Using antilog table for 0.8451: mantissa gives 7.00 approximately x ≈ 7000 </Solution> </ICSEExample>Solving Logarithmic Equations
<ICSEExample title="Solve log₂ x + log₂ 4 = 5"> <Solution> log₂ x + log₂ 4 = 5 log₂(4x) = 5 4x = 2⁵ = 32 x = 8 Verify: log₂ 8 + log₂ 4 = 3 + 2 = 5 ✓ </Solution> </ICSEExample> <ICSEExample title="Solve log x + log(x + 3) = 1"> <Solution> log x + log(x + 3) = 1 log[x(x + 3)] = 1 x(x + 3) = 10¹ = 10 x² + 3x — 10 = 0 (x + 5)(x — 2) = 0 x = -5 or x = 2But x must be positive (log of negative is undefined). Therefore, x = 2. </Solution> </ICSEExample>
Common Mistakes With Fixes
| Mistake | Correction |
|---|---|
| log(m + n) = log m + log n | Product rule: log(mn) = log m + log n (not sum) |
| log(m — n) = log m — log n | Quotient rule: log(m/n) = log m — log n (not difference) |
| Forgetting domain restrictions | Argument must be positive, base > 0 and ≠ 1 |
| Cancelling log without checking base | log f(x) = log g(x) implies f(x) = g(x) only if same base |
ICSE Exam Focus
| Topic | Marks (approx.) | Frequency |
|---|---|---|
| Applying laws of logarithms | 3-4 marks | Very common |
| Solving logarithmic equations | 4 marks | Common |
| Characteristic and Mantissa | 2-3 marks | Occasionally asked |
| Using log tables | 3 marks | Common |
Self-Test
Q1: Express in logarithmic form: 3⁴ = 81
Q2: Simplify: log₂ 32 + log₂ 8 — log₂ 4
Q3: Solve: log₁₀(2x + 1) — log₁₀(x — 2) = 1
Q4: If log₁₀ 2 = 0.3010 and log₁₀ 3 = 0.4771, find log₁₀ 6 and log₁₀ 1.5.
Q5: Solve: log₃(5x — 1) = 2
