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.