Indices and logarithms
An index says how many times a base is multiplied by itself. A logarithm is that index: 10³ is 1,000, so log 1,000 is 3. Logs turn multiplication into addition.
Laws of indices
| Law | Rule | Example |
|---|---|---|
| Product | aᵐ × aⁿ = a^(m+n) | 2³ × 2⁴ = 2⁷ = 128 |
| Quotient | aᵐ ÷ aⁿ = a^(m−n) | 5⁶ ÷ 5⁴ = 5² = 25 |
| Power of a power | (aᵐ)ⁿ = a^(mn) | (3²)³ = 3⁶ = 729 |
| Zero index | a⁰ = 1 | 7⁰ = 1 |
| Negative index | a^(−n) = 1/aⁿ | 2^(−3) = 1/8 |
| Fractional index | a^(m/n) = (ⁿ√a)ᵐ | 8^(2/3) = (∛8)² = 4 |
| Power of a product | (ab)ⁿ = aⁿbⁿ | (2x)³ = 8x³ |
Laws of logarithms
| Law | Rule | Example |
|---|---|---|
| Definition | log_b x = y means bʸ = x | log₂ 8 = 3 because 2³ = 8 |
| Product | log(xy) = log x + log y | log₂ 8 + log₂ 4 = log₂ 32 = 5 |
| Quotient | log(x/y) = log x − log y | log 1,000 − log 10 = log 100 = 2 |
| Power | log(xⁿ) = n log x | log₃ 81 = log₃ 3⁴ = 4 |
| Special values | log_b b = 1, log_b 1 = 0 | log 10 = 1, log 1 = 0 (log with no base written means base 10) |
| Change of base | log_b x = log x ÷ log b | log₂ 64 = log 64 ÷ log 2 = 6 |
Simplify with the index laws
(a) Simplify (2x³y²)³ ÷ (4x⁵y). (b) Evaluate 27^(2/3) × 16^(−1/2).
- (a) Power of a product: (2x³y²)³ = 2³ × x⁹ × y⁶ = 8x⁹y⁶.
- Divide: 8 ÷ 4 = 2; x⁹ ÷ x⁵ = x⁴; y⁶ ÷ y¹ = y⁵.
- (b) 27^(2/3) = (∛27)² = 3² = 9.
- 16^(−1/2) = 1 ÷ 16^(1/2) = 1 ÷ 4 = 1/4.
- Multiply: 9 × 1/4 = 9/4.
(a) 2x⁴y⁵ (b) 9/4 = 2.25
Solve index equations
Solve (a) 2^(x+1) = 32 (b) 3^(2x) = 81.
- (a) Write 32 as a power of 2: 32 = 2⁵.
- So 2^(x+1) = 2⁵. Same base, so the indices are equal: x + 1 = 5.
- x = 4. Check: 2⁵ = 32. ✓
- (b) 81 = 3⁴, so 3^(2x) = 3⁴ and 2x = 4.
- x = 2. Check: 3⁴ = 81. ✓
(a) x = 4 (b) x = 2
Logs answer 'how long?'
Rs. 100,000 is invested at 8% a year, compounded yearly. After how many years does it double?
- The amount after n years is 100,000 × 1.08ⁿ. We need 1.08ⁿ = 2.
- Take logs of both sides: n × log 1.08 = log 2.
- n = log 2 ÷ log 1.08 = 0.30103 ÷ 0.033424 = 9.01 years.
- Check: after 9 years, 100,000 × 1.08⁹ = Rs. 199,900.46, just under double.
- After 10 years it is Rs. 215,892.50.
About 9.01 years by the formula. With yearly compounding it is just short of double after 9 years and first passes Rs. 200,000 at the end of year 10.
A Sri Lankan bank officer uses logs to answer 'how many years until a fixed deposit doubles?': n = log 2 ÷ log(1 + r). The rate r always comes from the bank's current offer, since real rates change.
Short answer: “Evaluate 27^(2/3) × 9^(−1/2).” 27^(2/3) = 9 and 9^(−1/2) = 1/3, so the answer is 3.
Short answer: “Find log₂ 64.” 2⁶ = 64, so log₂ 64 = 6.
Structured question: “How many years for Rs. 100,000 to double at 8% compounded yearly?” 1.08ⁿ = 2; n = log 2 ÷ log 1.08 = 0.30103 ÷ 0.033424 ≈ 9.01 years.
Index laws: multiply → add the powers, divide → subtract, power of a power → multiply. Log laws copy them: log(xy) adds, log(x/y) subtracts, log(xⁿ) brings n down in front.