Operations on the Set of Real Numbers

Lessons

Senior 4 · Mathematics

Operations on the Set of Real Numbers

5 lessons

Lesson 5 of 5

Modelling with logarithms: growth, decay, interest and earthquakes

25 min

Change by a factor

Some quantities change by the same factor in each equal time step rather than by the same amount: a savings balance that earns compound interest, a population that grows by a fixed percentage, a radioactive substance that loses half its atoms in each half-life. Their models have the unknown in the exponent, and the decimal logarithm is the tool that brings it down. Earthquake magnitude works the other way round: the scale is a decimal logarithm. This lesson models all four situations.

Definition

Exponential model: a quantity $N$ that is multiplied by the same factor $a \gt 0$ in each unit of time follows

N(t) = N_0\, a^t ,

where $N_{0} = N (0)$ is the starting value. If $a \gt 1$ it is exponential growth; if $0 \lt a \lt 1$ , exponential decay. A growth rate $r$ per unit time gives $a = 1 + r$ ; a decay rate $r$ gives $a = 1 - r$ .

Doubling time (growth): the time for $N$ to double. Half-life (decay): the time for $N$ to halve.

Example: a 2.3% yearly growth rate gives $a = 1.023$ ; a 15% hourly decay rate gives $a = 0.85$ .

Finding the time

To find when the model reaches a value $N$ , divide by $N_{0}$ and take decimal logarithms:

N_0\, a^t = N \iff a^t = \frac{N}{N_0} \iff t = \frac{\log\frac{N}{N_0}}{\log a} .

Putting $\frac{N}{N_0} = 2$ gives the doubling time $T_2 = \frac{\log 2}{\log a}$ , and putting $\frac{N}{N_0} = \frac{1}{2}$ gives the half-life $T_{\frac{1}{2}} = \frac{\log 0.5}{\log a}$ . Neither depends on $N_{0}$ : a growing quantity takes the same time to go from 1 to 2 as from 2 to 4.

Worked example 1: population growth

A district has 420 000 inhabitants and grows at 2.3% a year (illustrative figures). Assuming the rate stays constant, (a) when will the population first exceed 500 000? (b) What is its doubling time?

Solution The model is $P(t) = 420\,000 \times 1.023^t$ , with $t$ in years.

$1.023^t = \frac{500\,000}{420\,000}$ , so $t = \frac{\log(1.1905)}{\log 1.023} = \frac{0.07572}{0.009876} \approx 7.67$ . The population first exceeds 500 000 during the 8th year. (Check: $P(8) \approx 503\,800$ .)
$T_2 = \frac{\log 2}{\log 1.023} = \frac{0.30103}{0.009876} \approx 30.5$ years.

Real populations do not keep a constant rate for decades, so a model like this is trusted only for short-term planning.

Worked example 2: radioactive decay

Iodine-131, used in hospitals to treat some thyroid diseases, has a half-life of about 8.02 days. (a) What percentage of a dose is left after 30 days? (b) After how many days does only 1% remain?

Solution Halving every 8.02 days gives $N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{8.02}}$ .

$\frac{N}{N_0} = 0.5^{\frac{30}{8.02}} = 0.5^{3.741} \approx 0.0748$ : about 7.5% remains.
$0.5^{\frac{t}{8.02}} = 0.01$ , so $\frac{t}{8.02}\log 0.5 = \log 0.01 = -2$ and $t = \frac{-2 \times 8.02}{-0.30103} \approx 53.3$ days.

Compound interest

A principal $P$ invested at a nominal annual rate $r$ , with interest added $k$ times a year, grows in $t$ years to

A = P\left(1 + \frac{r}{k}\right)^{kt} .

This is an exponential model with factor $1 + \frac{r}{k}$ per period. Logarithms answer the two harder questions. How long?

t = \frac{\log\frac{A}{P}}{k\log\left(1 + \frac{r}{k}\right)} .

What rate? Take the $k t$ -th root: $1 + \frac{r}{k} = \left(\frac{A}{P}\right)^{\frac{1}{kt}}$ , which a calculator evaluates as $10^{\frac{1}{kt}\log\frac{A}{P}}$ .

Worked example 3: a savings account at a SACCO

Uwase deposits 800 000 Frw in a savings and credit cooperative that pays 7.5% a year, compounded quarterly (an illustrative rate). (a) Find the balance after 3 years. (b) When does the balance first reach 1 200 000 Frw? (c) What annual rate, compounded monthly, would double her money in 8 years?

Solution The factor per quarter is $1 + \frac{0.075}{4} = 1.01875$ .

$A = 800\,000 \times 1.01875^{12} \approx 999\,773$ Frw.
$t = \frac{\log 1.5}{4\log 1.01875} = \frac{0.17609}{4 \times 0.0080676} \approx 5.46$ years, that is, after 21.8 quarters. Interest is added only at the end of a quarter, so the balance first reaches 1 200 000 Frw at the end of the 22nd quarter, after 5.5 years.
$\left(1 + \frac{r}{12}\right)^{96} = 2$ , so $1 + \frac{r}{12} = 2^{\frac{1}{96}} = 10^{\frac{\log 2}{96}} \approx 1.007246$ , and $r \approx 12 \times 0.007246 \approx 0.0870$ : about 8.70% a year.

The magnitude of an earthquake

A seismograph records the shaking of the ground as a wavy trace. Charles Richter defined the magnitude of an earthquake from the largest amplitude $A$ on a standard instrument, compared with a fixed reference amplitude $A_{0}$ :

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

So the magnitude is the power of ten by which the shaking beats the reference, and the US Geological Survey (USGS) states the consequence plainly: each whole-number increase in magnitude means a tenfold increase in measured amplitude and about 32 times more energy released. Since $10^{1.5} \approx 31.6$ , the energy rule is $E \propto 10^{1.5M}$ . For two earthquakes,

\frac{A_1}{A_2} = 10^{M_1 - M_2}, \qquad \frac{E_1}{E_2} = 10^{1.5(M_1 - M_2)} .

Difference in magnitude	1	2	3
Amplitude ratio	10	100	1000
Energy ratio (about)	32	1000	32 000

Seismologists now quote the closely related moment magnitude $M_{w}$ , which keeps the same logarithmic steps.

Worked example 4: the 2008 Lake Kivu earthquake

On 3 February 2008 an earthquake of moment magnitude 5.9 struck the southern shore of Lake Kivu, near Bukavu (DR Congo) and Cyangugu (Rwanda), and killed at least 39 people. Nine of its aftershocks had magnitudes above 3.7. Treating both on the same scale, compare the main shock with a magnitude 3.7 aftershock. Then find the magnitude of a tremor whose amplitude is 5000 times $A_{0}$ .

Solution

Amplitude ratio: $10^{5.9 - 3.7} = 10^{2.2} \approx 158$ .
Energy ratio: $10^{1.5 \times 2.2} = 10^{3.3} \approx 2000$ . The main shock released about 2000 times the energy of the aftershock.
$M = \log 5000 = \log 5 + 3 \approx 3.7$ .

The logarithm makes a huge range of earthquakes fit on a short scale, but it also hides how large the jumps are: 5.9 "only" looks 2.2 more than 3.7.

Tip

Every problem in this lesson ends with the same move: isolate the power, $a^t = \text{number}$ , then take logarithms of both sides. Write the model first, in words and symbols, before you calculate, and ask at the end whether the answer is sensible (a time cannot be negative; interest is only added at the end of a period).

IRIS Earthquake Science: Earthquake Magnitude, using pasta to understand magnitudes. Watch for how many strands of spaghetti stand for one magnitude step, and why the energy grows about 32 times per step.

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Practice

A calculator is allowed. Give times to 3 significant figures unless told otherwise.

A model is $N(t) = 250 \times 1.08^t$ . Find (a) $N (5)$ (b) the doubling time.
A car loses 12% of its value each year. How long until it is worth half its new price?
A town of 60 000 grows at 3% a year (illustrative). After how many whole years will it first exceed 90 000?
Carbon-14 has a half-life of about 5730 years. What fraction of the carbon-14 in a piece of wood is left after 2000 years?
500 000 Frw is invested at 6% a year compounded monthly. Find the balance after 4 years, to the nearest franc.
How long does it take money to triple at 9% a year compounded yearly? Give the answer in whole years, and explain.
Earthquake P has magnitude 6.4 and earthquake Q magnitude 4.9. How many times larger is P's amplitude, and about how many times more energy did it release?
A seismograph records an amplitude 80 000 times the reference amplitude. Find the magnitude to 1 decimal place.

Show answersHide answers

(a) $250 \times 1.08^5 \approx 367.3$ (b) $\frac{\log 2}{\log 1.08} \approx 9.01$ .
${0.88}^{t} = 0.5$ , so $t = \frac{\log 0.5}{\log 0.88} \approx 5.42$ years.
$t = \frac{\log 1.5}{\log 1.03} \approx 13.7$ , so after 14 years. (Check: $60\,000 \times 1.03^{14} \approx 90\,755$ .)
$0.5^{\frac{2000}{5730}} \approx 0.785$ : about 78.5% remains.
$500\,000 \times 1.005^{48} \approx 635\,245$ Frw.
${1.09}^{t} = 3$ gives $t = \frac{\log 3}{\log 1.09} \approx 12.7$ . Interest is added once a year, so the money has tripled after 13 years.
Amplitude: $10^{1.5} \approx 31.6$ times. Energy: $10^{2.25} \approx 178$ times.
$M = \log 80\,000 \approx 4.9$ .

Quick check

A quantity follows N = 500 × 0.9ᵗ. Which statement is true?

Key points

$N = N_{0} a^{t}$ : growth if $a \gt 1$ , decay if $0 \lt a \lt 1$ ; $t = \frac{\log\frac{N}{N_0}}{\log a}$ .
Doubling time $\frac{\log 2}{\log a}$ and half-life $\frac{\log 0.5}{\log a}$ do not depend on $N_{0}$ .
$A = P\left(1 + \frac{r}{k}\right)^{kt}$ ; interpret the time in whole periods.
$M = \log\frac{A}{A_0}$ : one magnitude step is 10 times the amplitude and about 32 times the energy.

1Change by a factor

2Finding the time

3Compound interest

4The magnitude of an earthquake

Change by a factor

Finding the time

Compound interest

The magnitude of an earthquake