Mathematics Advanced • Year 11 • Module 4 • Lesson 7
Laws of Logarithms
Apply the product, quotient and power laws to simplify exam expressions and to derive useful facts (slide-rule arithmetic, decibel addition, scientific notation).
Problem 1 — Slide rule arithmetic (engineering)
Before calculators, engineers multiplied numbers by adding log10 values from a table. Use log102 ≈ 0.3010 and log103 ≈ 0.4771 throughout.
Set up: What are we solving for?
(i) Use the product law to compute log106 to 4 decimal places. 1 mark
(ii) Hence find log10(210) and use it to estimate 210 without computing 2 · 2 · … · 2. 3 marks
(iii) Use the laws to compute log1072. (Hint: 72 = 8 × 9.) 2 marks
Stuck? Revisit lesson § Concept — the laws turn multiplication into addition.Problem 2 — Adding sound sources (decibels)
The decibel level of two independent sound sources combines via
Ltotal = 10 log10(I1/I0 + I2/I0)
Set up: What are we solving for?
(i) Two identical lawnmowers each register 90 dB. Show that the combined level is Ltotal = 90 + 10 log10(2). Then evaluate Ltotal using log102 ≈ 0.3010. 3 marks
(ii) Show that doubling the number of identical sources adds 10 log102 ≈ 3.01 dB, regardless of the starting level. 2 marks
(iii) Explain in one sentence why "two 90-dB sources are not twice as loud as one" — what does the log law tell us about the perceived doubling? 2 marks
Problem 3 — Scientific-notation magnitudes
For a positive number N expressed in scientific notation as N = m × 10k (with 1 ≤ m < 10 and integer k),
log10N = log10m + k
Set up: What are we solving for?
(i) Using the product law, derive the formula above from N = m × 10k. 2 marks
(ii) The mass of the Earth is ≈ 5.972 × 1024 kg. Compute log10(mass) using log105.972 ≈ 0.776. 2 marks
(iii) A bacterium has mass ≈ 10−15 kg. By how many orders of magnitude does the Earth's mass exceed the bacterium's? Justify using the difference of logs. 2 marks
Stuck? The "order of magnitude" between two numbers is exactly log10(ratio).Problem 4 — Earthquake amplitude ratios
For two earthquakes with amplitudes A1 and A2, the magnitude difference is
M1 − M2 = log10(A1/A2)
Set up: What are we solving for?
(i) Use the quotient law to derive the formula above from M(A) = log10(A/A0). 2 marks
(ii) The 2011 Tohoku earthquake was magnitude 9.1; a moderate aftershock was magnitude 6.1. Find the amplitude ratio Amain/Aaftershock. 2 marks
(iii) Energy released scales as E ∝ A3/2. Using the power law, express the magnitude difference in terms of the energy ratio E1/E2. Hence find the energy ratio for the two earthquakes in (ii). 3 marks
Problem 5 — Chemistry: pH + pOH = 14
For an aqueous solution at 25°C, the hydrogen-ion and hydroxide-ion concentrations are related by
[H+] · [OH−] = Kw = 10−14, with pH = −log10[H+], pOH = −log10[OH−]
Set up: What are we solving for?
(i) Use the product law to prove that pH + pOH = 14. 3 marks
(ii) Household ammonia has pH ≈ 11.6. Find pOH and the [OH−] concentration. 2 marks
(iii) Explain in one sentence why diluting an acidic solution by a factor of 10 raises its pH by exactly 1. 2 marks
Stuck on (iii)? Diluting by 10 divides [H+] by 10; use the quotient/power law.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Slide rule arithmetic
Set up. We are using log10(MN) = log10M + log10N (and the power law) to convert multiplications/exponents into additions.
(i) log106 = log10(2 × 3) = log102 + log103 ≈ 0.3010 + 0.4771 = 0.7781.
(ii) log10(210) = 10 log102 ≈ 10 × 0.3010 = 3.010. So 210 ≈ 103.010 = 103 × 100.010. Since 100.010 ≈ 1.023, 210 ≈ 1023. (Actual value: 1024 — error < 0.1%.)
(iii) log1072 = log10(8 × 9) = log108 + log109 = 3 log102 + 2 log103 ≈ 3(0.3010) + 2(0.4771) = 0.9030 + 0.9542 = 1.8572.
Problem 2 — Decibel addition
Set up. We are using log10(MN) = log10M + log10N to show how doubling intensity adds a fixed number of dB.
(i) Each mower has I/I0 = 109. Sum: 2 × 109. So Ltotal = 10 log10(2 × 109) = 10[log102 + 9] = 10 log102 + 90 = 90 + 10 log102 ≈ 90 + 3.01 = 93.01 dB.
(ii) Doubling the source count doubles the total intensity. Lnew = 10 log10(2I/I0) = 10[log102 + log10(I/I0)] = 10 log102 + Lold. Difference = 10 log102 ≈ 3.01 dB, independent of the starting level.
(iii) The decibel scale is logarithmic: doubling the intensity adds only ≈ 3 dB, but a "perceived doubling of loudness" needs roughly +10 dB (a tenfold intensity increase). The log law translates physical doubling into a small additive change in dB, mirroring how the ear actually perceives loudness.
Problem 3 — Scientific notation
Set up. We use the product and power laws to read off the "order of magnitude" of a number directly from its scientific notation.
(i) log10N = log10(m × 10k) = log10m + log10(10k) [product law] = log10m + k log1010 [power law] = log10m + k. ✓
(ii) log10(5.972 × 1024) = log105.972 + 24 ≈ 0.776 + 24 = 24.776.
(iii) log10(MEarth/Mbact) = log10MEarth − log10Mbact = 24.776 − (−15) = 39.776 ≈ 40 orders of magnitude. The Earth is about 1040 times more massive than a single bacterium.
Problem 4 — Earthquake ratios
Set up. We use the quotient and power laws to convert magnitude differences into multiplicative amplitude and energy ratios.
(i) M1 − M2 = log10(A1/A0) − log10(A2/A0) = log10[(A1/A0) ÷ (A2/A0)] = log10(A1/A2). ✓
(ii) 9.1 − 6.1 = 3.0 = log10(Amain/Aaft), so Amain/Aaft = 103 = 1000.
(iii) M1 − M2 = log10(A1/A2) and E ∝ A3/2, so A1/A2 = (E1/E2)2/3. Take log10: M1 − M2 = (2/3) log10(E1/E2). So log10(E1/E2) = (3/2)(M1 − M2) = (3/2)(3.0) = 4.5. Emain/Eaft = 104.5 ≈ 31 600. The mainshock released about 32 000 times more energy than the aftershock.
Problem 5 — pH + pOH = 14
Set up. We use the product law to convert a multiplicative relationship between concentrations into an additive relationship between pH and pOH.
(i) Take log10 of [H+] · [OH−] = 10−14:
log10[H+] + log10[OH−] = −14 (product law).
Multiply by −1: −log10[H+] − log10[OH−] = 14, i.e. pH + pOH = 14. ▮
(ii) pOH = 14 − 11.6 = 2.4. [OH−] = 10−2.4 ≈ 4.0 × 10−3 mol/L.
(iii) If [H+] is divided by 10, then log10[H+] decreases by 1 (quotient law: log10(c/10) = log10c − 1). Since pH = −log10[H+], pH increases by exactly 1 — a unit step in pH always corresponds to a tenfold dilution.