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Indices and Logarithms

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Indices and Logarithms

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Indices (exponents) and logarithms are closely linked topics that frequently appear together in WASSCE Paper 2. Indices deal with powers — for example, 2³ = 8 means 2 raised to the power 3 equals 8. Logarithms are the inverse operation: log₂ 8 = 3 because 2³ = 8. In general, if a^x = N, then log_a N = x.

Essential Formulas:

  • a^m × a^n = a^(m+n)
  • a^m ÷ a^n = a^(m−n)
  • (a^m)^n = a^(mn)
  • a^0 = 1 (for any a ≠ 0)
  • a^(−n) = 1/a^n
  • a^(m/n) = (ⁿ√a)^m = ⁿ√(a^m)
  • log_a (MN) = log_a M + log_a N
  • log_a (M/N) = log_a M − log_a N
  • log_a (M^n) = n log_a M
  • Change of base: log_a M = log_b M / log_b a

Key Facts:

  • log_a 1 = 0 for any base a (since a^0 = 1)
  • log_a a = 1 (since a^1 = a)
  • If log_a x = log_a y, then x = y (one-to-one property)
  • When solving exponential equations, try to express both sides with the same base if possible
  • Common logarithm base is 10: log x means log₁₀ x. Natural logarithm base is e ≈ 2.71828: ln x means log_e x

⚡ Exam Tip: WASSCE often asks you to solve equations like 2^(x+1) = 32. Write 32 as 2^5, then equate exponents: x + 1 = 5, so x = 4. When solving log equations, always check that the argument is positive — a log equation like log₂(x − 3) = 4 requires x − 3 > 0, giving x > 3. Substitute x = 11 into the original: log₂(8) = 3 ≠ 4. Wait, let me recalculate: x − 3 = 2^4 = 16, so x = 19. Always verify your solution in the original equation.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Laws of Indices

The laws of indices apply to all real exponents (integers, fractions, negatives). They are fundamental to simplifying expressions and solving equations involving powers.

Product law: a^m × a^n = a^(m+n). Example: 3² × 3⁴ = 3^(2+4) = 3⁶ = 729. Quotient law: a^m ÷ a^n = a^(m−n). Example: 5⁶ ÷ 5² = 5^(6−2) = 5⁴ = 625. Power law: (a^m)^n = a^(mn). Example: (2³)⁴ = 2^(3×4) = 2¹² = 4096.

Negative indices indicate reciprocals: 2^(−3) = 1/2³ = 1/8. Fractional indices indicate roots: 8^(2/3) = (³√8)² = 2² = 4.

Solving Exponential Equations

An exponential equation has the unknown in the exponent. Two approaches:

Same base method: If both sides can be written with the same base, equate the exponents. Example: 3^(2x+1) = 27 = 3³ → 2x + 1 = 3 → x = 1.

Logarithm method: If same base is not possible, take logarithms of both sides. Example: 2^x = 15 → x log 2 = log 15 → x = log 15 / log 2 ≈ 3.907.

Understanding Logarithms

The logarithm of a number is the exponent to which the base must be raised to produce that number. Formally: if a^x = N (a > 0, a ≠ 1), then log_a N = x.

Common logarithm (base 10): log₁₀ x is usually written as log x. log 100 = 2 because 10² = 100. Natural logarithm (base e): log_e x is written as ln x. e ≈ 2.71828. Any base: log₂ 8 = 3 because 2³ = 8.

Logarithmic Laws

These mirror the index laws since logarithms and indices are inverse operations:

  • log(MN) = log M + log N
  • log(M/N) = log M − log N
  • log(M^n) = n log M
  • log_a a = 1; log_a 1 = 0

Solving Logarithmic Equations

Example: log₂(x + 3) + log₂(x − 1) = 3 Combine logs: log₂[(x + 3)(x − 1)] = 3 Convert to exponential: (x + 3)(x − 1) = 2³ = 8 Expand: x² + 3x − x − 3 = 8 → x² + 2x − 11 = 0 Solve: x = (−2 ± √(4 + 44)) / 2 = (−2 ± √48) / 2 = (−2 ± 4√3) / 2 = −1 ± 2√3 Check domain: x + 3 > 0 and x − 1 > 0 → x > 1. Both solutions are > 1? −1 − 2√3 < 0, reject. x = −1 + 2√3 ≈ 2.46, valid.

Change of Base Formula

log_a M = log_b M / log_b a. This allows you to compute logs of any base using a calculator that only has common or natural log functions.

Example: Calculate log₂ 7 using a calculator. log₂ 7 = log₁₀ 7 / log₁₀ 2 ≈ 0.8451 / 0.3010 ≈ 2.807.

Problem-Solving Strategies:

  • When given an equation like 4^(x+1) = 8^(2x), rewrite both sides with base 2: (2²)^(x+1) = (2³)^(2x) → 2^(2x+2) = 2^(6x) → 2x + 2 = 6x → x = 0.5
  • For equations combining indices and logarithms, isolate the logarithmic or exponential term first
  • Graphs of y = a^x and y = log_a x are inverses — symmetric about y = x

Common Mistakes:

  • Confusing a^(mn) with a^m × a^n: (2³)² = 2^6 = 64, but 2³ × 2² = 8 × 4 = 32 — different results!
  • Forgetting that log(M + N) ≠ log M + log N — only products factor, not sums
  • Applying log to negative numbers — arguments of logarithms must always be positive
  • Mixing up log laws when expanding: log(x²/9) = log x² − log 9 = 2 log x − log 9, not 2 log x − 3

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Derivation of Logarithmic Laws

Product law: Let log_a M = x and log_a N = y, so M = a^x and N = a^y. Then MN = a^x × a^y = a^(x+y). Taking log base a: log_a (MN) = x + y = log_a M + log_a N. ✓

Quotient law: MN = a^x / a^y = a^(x−y). So log_a (M/N) = x − y = log_a M − log_a N. ✓

Power law: M = a^x, so M^n = (a^x)^n = a^(nx). Therefore log_a (M^n) = nx = n log_a M. ✓

The Number e and Natural Logarithms

The constant e = 2.71828… is defined as the limit of (1 + 1/n)^n as n → ∞. It arises naturally in compound interest, population growth, and radioactive decay. The function e^x has the unique property that its derivative equals itself: d/dx(e^x) = e^x.

The natural logarithm ln x = log_e x has the derivative: d/dx(ln x) = 1/x. This makes it indispensable in calculus.

Exponential and Logarithmic Graphs

y = a^x: passes through (0, 1), always positive, increases for a > 1, decreases for 0 < a < 1, has a horizontal asymptote at y = 0.

y = log_a x: passes through (1, 0), domain x > 0, range all real numbers, increases for a > 1, has a vertical asymptote at x = 0.

These two graphs are reflections of each other across the line y = x.

Solving Systems Involving Logarithms

Example: Given that log₂(x + y) = 5 and log₂(x − y) = 3, find x and y. From log₂(x + y) = 5: x + y = 2⁵ = 32 From log₂(x − y) = 3: x − y = 2³ = 8 Adding: 2x = 40 → x = 20. Then y = 12.

Application: pH and Decibel Calculations

pH = −log₁₀[H⁺]. A solution with [H⁺] = 1.5 × 10⁻⁵ mol/L has pH = −log(1.5 × 10⁻⁵) = −(log 1.5 + log 10⁻⁵) = −(0.176 + (−5)) = 4.824.

Sound level in decibels: L = 10 log₁₀(I/I₀), where I₀ = 10⁻¹² W/m².

Simultaneous Logarithmic and Exponential Equations

Example: Given y = 2^x and log₂ y + log₂ x = 3 Substitute: y = 2^x gives log₂(2^x) + log₂ x = 3 → x + log₂ x = 3 This requires numerical solution or inspection: try x = 2: 2 + 1 = 3 ✓. So x = 2, y = 4.

WASSCE Exam Patterns

Paper 1: Index questions often appear as simplification problems, evaluating expressions like (8/27)^(2/3) or solving 4^x = 0.125. Logarithm questions frequently test the laws directly: “Express log₃ 54 − log₃ 2 as a single logarithm.”

Paper 2: Solving exponential equations leading to quadratic equations. Example: 4^x − 3(2^x) + 2 = 0 can be written as (2^x)² − 3(2^x) + 2 = 0. Let u = 2^x, then u² − 3u + 2 = 0 → (u − 1)(u − 2) = 0 → u = 1 or u = 2. So 2^x = 1 → x = 0, or 2^x = 2 → x = 1.

⚡ Advanced Tip: When solving 3^(2x+1) − 8(3^x) = 3, treat it as a quadratic in 3^x. Let u = 3^x, then 3(u²) − 8u − 3 = 0 (since 3^(2x+1) = 3 × (3^x)² = 3u²). Solve: 3u² − 8u − 3 = 0 → (3u + 1)(u − 3) = 0 → u = 3 or u = −1/3. Since u = 3^x > 0, u = 3. So 3^x = 3 → x = 1.

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