Logarithmic and Exponential Equations

Lessons

Senior 5 · Mathematics

Logarithmic and Exponential Equations

4 lessons

Lesson 4 of 4

Growth, decay, carbon dating and earthquakes

25 min

Change by a fixed percentage

When a quantity changes by the same percentage in each equal time interval, it follows an exponential model, and logarithms answer the question "how long?". This lesson applies the model to four things: a growing town, a radioactive medicine, an ancient piece of charcoal and an earthquake. The last one goes further: the magnitude of an earthquake is itself a logarithm, and so is the planning question "how likely is a large earthquake in the next 30 years?".

Definition

Exponential growth or decay model: $Q(t) = Q_0\,b^t$ , where $Q_{0} = Q (0)$ is the starting value and $b \gt 0$ is the factor per unit of time. If the quantity grows by $r$ (as a decimal) each unit, $b = 1 + r$ ; if it shrinks by $r$ , $b = 1 - r$ .

Half-life $T$ of a radioactive substance: the time in which half of any sample decays. Then $Q(t) = Q_0\left(\frac{1}{2}\right)^{\frac{t}{T}}$ .

Example: a town of 85,000 people growing by 3.2% a year has $P(t) = 85\,000 \times 1.032^t$ after $t$ years.

Worked example 1: a growing town

A town has 85,000 people and grows by 3.2% a year (illustrative figures). (a) Find its population after 8 years. (b) After how many years will it pass 150,000? (c) Another town grew from 60,000 to 75,000 in 6 years. What was its yearly growth rate?

Solution

$P(8) = 85\,000 \times 1.032^8 \approx 109\,359$ people.
$1.032^t = \frac{150\,000}{85\,000}$ , so $t = \frac{\log(150 \div 85)}{\log 1.032} = \frac{0.24667}{0.01368} = 18.03$ . It passes 150,000 a little after 18 years.
$60\,000\,(1 + r)^6 = 75\,000$ , so $1 + r = 1.25^{\frac{1}{6}} = 1.0379$ : about 3.8% a year.

Radioactive decay

Iodine-131 is used in hospitals to diagnose and treat thyroid disease. Its half-life is 8.02 days, so a dose loses half its activity every 8.02 days, whatever its size (Figure 1). To find a time, take logarithms of $\left(\frac{1}{2}\right)^{\frac{t}{T}} = \frac{Q}{Q_0}$ :

t = T\,\frac{\log\frac{Q}{Q_0}}{\log\frac{1}{2}}.

Figure 1. Radioactive decay of iodine-131 (half-life 8.02 days): equal times give equal fractions, not equal amounts.

Worked example 2: a medical isotope

A hospital receives iodine-131 with an activity of 200 MBq (megabecquerels). (a) Find its activity after 30 days. (b) How long until it falls to 5% of its first value?

Solution

$A = 200 \times 0.5^{\frac{30}{8.02}} = 200 \times 0.0748 = 15.0$ MBq.
$0.5^{\frac{t}{8.02}} = 0.05$ , so $t = 8.02 \times \frac{\log 0.05}{\log 0.5} = 8.02 \times 4.322 = 34.7$ days.

Carbon dating

Living plants and animals take in carbon, including a small, steady share of radioactive carbon-14. When they die, the intake stops and the carbon-14 decays with a half-life of about 5730 years. Measuring the fraction that is left in wood, charcoal or bone gives the time since death. The method works for remains up to about 50,000 years old; after that too little carbon-14 is left to measure.

Worked example 3: dating charcoal

Charcoal from an old furnace contains 79% of the carbon-14 found in living wood (illustrative figure). Estimate its age.

Solution

\left(\frac{1}{2}\right)^{\frac{t}{5730}} = 0.79 \quad \Rightarrow \quad t = 5730 \times \frac{\log 0.79}{\log 0.5} = 5730 \times 0.3401 \approx 1950 \text{ years}.

A sensible answer: less than one half-life, since more than half the carbon-14 is left.

Earthquake magnitude

The Richter (local) magnitude of an earthquake is a logarithm. If a standard seismograph 100 km from the centre records a largest wave of amplitude $A$ , and $A_{0}$ is a fixed reference amplitude, then

M = \log\frac{A}{A_0}.

Each step of 1 in magnitude means 10 times the amplitude. The energy released grows faster: approximately $\log E = 4.8 + 1.5M$ ( $E$ in joules), so each step of 1 multiplies the energy by $10^{1.5} \approx 31.6$ . Scientists now usually quote the closely related moment magnitude, but the logarithmic idea is the same.

Worked example 4: comparing two earthquakes

Earthquake P has magnitude 5.8 and earthquake Q has magnitude 4.3. (a) How many times larger is P's recorded amplitude? (b) How many times more energy did P release? (c) A seismograph records an amplitude 2000 times $A_{0}$ . What is the magnitude?

Solution

$\frac{A_P}{A_Q} = \frac{A_0 \times 10^{5.8}}{A_0 \times 10^{4.3}} = 10^{1.5} \approx 31.6$ .
$\frac{E_P}{E_Q} = 10^{1.5 \times (5.8 - 4.3)} = 10^{2.25} \approx 178$ .
$M = \log 2000 = 3.3$ .

Earthquake risk

In a given region, small earthquakes are common and large ones rare. Seismologists summarise this with the Gutenberg–Richter law $\log N = a - bM$ , where $N$ is the average number of earthquakes per year with magnitude at least $M$ , and $a$ and $b$ are found from the region's records ( $b$ is usually close to 1). If $p$ is the probability of at least one such earthquake in a year, and years are independent, then the probability of at least one in $n$ years is

P_{n} = 1 - (1 - p)^{n} .

Engineers use numbers like these to decide how strongly to build schools, bridges and dams.

Figure 2. An illustrative Gutenberg–Richter law,

\log N = 4 - M

. On a logarithmic scale the relation is a straight line: each extra unit of magnitude makes earthquakes ten times rarer.

Worked example 5: the chance of a large earthquake

In a region with $\log N = 4 - M$ (illustrative values), take the yearly probability of an earthquake of magnitude 5 or more to be $p = 0.1$ . (a) Check $N$ for $M = 5$ . (b) Find the probability of at least one such earthquake in 30 years. (c) After how many years does that probability first exceed 50%?

Solution

$\log N = 4 - 5 = -1$ , so $N = 0.1$ : on average one every 10 years.
$P_{30} = 1 - 0.9^{30} = 1 - 0.042 = 0.958$ : almost certain.
$1 - 0.9^n \gt 0.5 \iff 0.9^n \lt 0.5 \iff n \gt \frac{\log 0.5}{\log 0.9} = 6.58$ (the sign turns because $\log 0.9 \lt 0$ ). So after 7 years.

Warning

A magnitude is not a measure you can add. Two earthquakes of magnitude 5 do not make a magnitude-10 event: the energy only doubles, which adds $\frac{\log 2}{1.5} \approx 0.2$ to the magnitude. And "once in 10 years on average" does not mean "exactly every 10 years": the probability of none at all in 10 years is still $0.9^{10} \approx 0.35$ .

Tip

Check the size of the answer. A decay time to fall below half must exceed one half-life; a fraction above $\frac{1}{2}$ must give an age below one half-life. Rough estimates like these catch most calculator slips.

Khan Academy: Richter scale. Watch how the scale compares earthquakes through the logarithm of the wave amplitude, so that one unit of magnitude means ten times the amplitude.

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Practice

Data in questions 1, 7 and 8 are illustrative. The last two are harder.

A district has 50,000 cattle, and the number grows by 2.5% a year. Find the number after 10 years, and the time for it to double.
Strontium-90 has a half-life of 28.8 years. How much of a 100 g sample is left after 50 years? How long until only 1 g is left?
A sample keeps 80% of its activity after 3 days. Find its half-life.
A bone contains 35% of the carbon-14 of living bone. Estimate its age.
Compare an earthquake of magnitude 6.2 with one of magnitude 5.0: how many times the amplitude, and how many times the energy?
Using $\log E = 4.8 + 1.5M$ , find the magnitude of an earthquake that releases $10^{15}$ J.
The yearly probability of a damaging flood in a valley is 0.04. Find the probability of at least one in 25 years, and the number of years after which that probability first exceeds 90%.
In a region, $\log N = 5 - M$ . Find $N$ for $M \ge 4$ and $M \ge 6$ , and explain what the two numbers mean.

Show answersHide answers

$50\,000 \times 1.025^{10} \approx 64\,004$ ; doubling time $\frac{\log 2}{\log 1.025} = 28.1$ years.
$100 \times 0.5^{\frac{50}{28.8}} = 30.0$ g; $t = 28.8 \times \frac{\log 0.01}{\log 0.5} = 191$ years.
$0.5^{\frac{3}{T}} = 0.8$ , so $T = \frac{3\log 0.5}{\log 0.8} = 9.32$ days.
$t = 5730 \times \frac{\log 0.35}{\log 0.5} \approx 8680$ years.
Amplitude $10^{1.2} \approx 15.8$ times; energy $10^{1.8} \approx 63.1$ times.
$15 = 4.8 + 1.5 M$ , so $M = 6.8$ .
$1 - 0.96^{25} = 0.640$ ; $0.96^n \lt 0.1$ gives $n \gt \frac{\log 0.1}{\log 0.96} = 56.4$ , so 57 years.
$N (4) = 10$ : about ten earthquakes of magnitude 4 or more each year. $N (6) = 0.1$ : about one of magnitude 6 or more every ten years.

1Change by a fixed percentage

2Radioactive decay

3Carbon dating

4Earthquake magnitude

5Earthquake risk

Change by a fixed percentage

Radioactive decay

Carbon dating

Earthquake magnitude

Earthquake risk