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Beyond: Growth and decay

Ratio and rates · step 5 of 5

A question to start

You put 1000 dollars in a savings account that pays 5% interest a year, and leave it alone for 30 years. One friend says you will have \(1000 + 30 \times 50 = 2500\) dollars. Another says it will be over 4300 dollars. Who is right? It depends on one small detail: whether the interest itself earns interest. That detail, applied year after year, is the difference between a straight line and an exponential curve, and it describes savings, debts, populations, medicines and radioactive fuel alike.

The idea

Simple and compound interest

The amount you start with is the principal \(P\). With simple interest you earn the same amount each year, a fixed percent of the principal. After \(t\) years at rate \(r\) (written as a decimal, so 5% is \(r = 0.05\)):

\[A = P(1 + rt).\]

That is linear growth: the same amount is added each year. With compound interest each year's interest is added to the account and earns interest too. Each year the amount is multiplied by \(1 + r\), the multiplier from Master:

\[A = P(1 + r)^t.\]

For the question, simple interest gives \(1000(1 + 0.05 \times 30) = 2500\) dollars, while compound interest gives \(1000 \times 1.05^{30} \approx 4321.94\) dollars. Both friends were right, about different accounts.

Compounding more often

Banks often add interest monthly or daily. If a yearly rate \(r\) is compounded \(n\) times a year, each period uses the rate \(\tfrac{r}{n}\), and there are \(nt\) periods:

\[A = P\left(1 + \frac{r}{n}\right)^{nt}.\]

More frequent compounding gives a little more, but not without limit. Take \(P = 1\), \(r = 1\) (100%) and \(t = 1\):

\(n\) 1 2 12 365 1 000 000
\(\left(1 + \frac1n\right)^n\) 2 2.25 2.613 2.715 2.718

The values close in on the number \(e \approx 2.71828\). Compounding "every instant", called continuous compounding, gives

\[A = Pe^{rt}.\]

Exponential growth and decay

All of these have the same shape: a starting amount times a fixed multiplier raised to the time.

\[N(t) = N_0 \, b^{\,t}\]

where \(N_0\) is the amount at time 0. If \(b > 1\) it is exponential growth; if \(0 < b < 1\) it is exponential decay. A quantity that shrinks by 15% a year has \(b = 0.85\).

Half-life and doubling time

For decay, it is natural to ask how long it takes to fall to half. That time, the half-life \(h\), is the same whatever amount you start with. After one half-life, \(\tfrac12\) is left; after two, \(\tfrac14\); after three, \(\tfrac18\):

\[N(t) = N_0 \left(\frac12\right)^{t/h}.\]

For growth, the matching idea is the doubling time. Growth at \(r\) per period doubles when \((1 + r)^t = 2\), and taking logarithms (see algebra, Beyond step):

\[t = \frac{\ln 2}{\ln(1 + r)}.\]

A handy estimate is the rule of 72: at \(R\%\) growth per period, the doubling time is about \(72 \div R\) periods. It works because \(\ln 2 \approx 0.693\) and, for small \(r\), \(\ln(1 + r) \approx r\); 72 is used instead of 69.3 because it is close and has so many factors, which makes the mental division easy.

Linear against exponential

A quantity growing by a fixed amount grows linearly; one growing by a fixed percent grows exponentially. In the short run they can look alike. In the long run exponential growth overtakes any linear growth, however slow the percent and however fast the line.

Worked examples

1. Carbon dating. Carbon-14 has a half-life of about 5730 years. A piece of old wood has 25% of the carbon-14 that a living tree has. How old is it?

  • 25% is \(\tfrac14 = \left(\tfrac12\right)^2\): two half-lives.
  • Age \(\approx 2 \times 5730 = 11\,460\) years.

If only 10% were left: \(\left(\tfrac12\right)^{t/5730} = 0.1\). Take logs: \(\dfrac{t}{5730}\ln 0.5 = \ln 0.1\), so \(t = 5730 \times \dfrac{\ln 0.1}{\ln 0.5} \approx 5730 \times 3.32 \approx 19\,000\) years.

2. Doubling time. At 7% a year, how long does money take to double?

  • Exactly: \(t = \dfrac{\ln 2}{\ln 1.07} \approx \dfrac{0.6931}{0.0677} \approx 10.2\) years.
  • Rule of 72: \(72 \div 7 \approx 10.3\) years. Very close.

3. A rover's power source. The Curiosity rover on Mars is powered by the heat from decaying plutonium-238, which has a half-life of about 87.7 years. What fraction of the fuel's original heat output is left after 14 years?

  • \(\left(\tfrac12\right)^{14/87.7} = 0.5^{0.1596} \approx 0.895\).
  • About 90% of the heat is still being produced. (The electricity the rover gets falls a little faster than this, because the parts that turn heat into electricity also wear out slowly; the 90% is for the fuel alone.)

Try it

  1. 2000 dollars is invested at 3% a year, compounded yearly. How much is there after 10 years?
  2. The same, but compounded monthly.
  3. The same, but compounded continuously.
  4. A medicine has a half-life of 6 hours in the body. A patient takes 200 mg. How much is left after 24 hours?
  5. A town of 5000 people grows by 4% a year. After how many years does it first pass 10 000?
  6. A car loses 15% of its value every year. After how many whole years is it worth less than half of what it cost?
Answers
  1. \(2000 \times 1.03^{10} \approx 2687.83\) dollars.
  2. \(2000 \left(1 + \dfrac{0.03}{12}\right)^{120} = 2000 \times 1.0025^{120} \approx 2698.71\) dollars.
  3. \(2000\,e^{0.03 \times 10} = 2000\,e^{0.3} \approx 2699.72\) dollars. Only about a dollar more than monthly.
  4. 24 hours is \(24 \div 6 = 4\) half-lives, so \(200 \times \left(\tfrac12\right)^4 = 200 \div 16 = 12.5\) mg.
  5. Doubling: \(t = \dfrac{\ln 2}{\ln 1.04} \approx 17.7\) years. After 17 years it is still just under 10 000 (\(5000 \times 1.04^{17} \approx 9740\)); after 18 years it is about 10 129. So 18 years. (Rule of 72: \(72 \div 4 = 18\).)
  6. Need \(0.85^t < 0.5\), so \(t > \dfrac{\ln 0.5}{\ln 0.85} \approx 4.27\). After 4 years it is worth \(0.85^4 \approx 52\%\); after 5 years, about 44%. So 5 years.

Common mistake

Adding the percents instead of multiplying the multipliers. 5% a year for 30 years is not 150% growth; it is \(1.05^{30} \approx 4.32\), a 332% increase. And a 50% loss followed by a 50% gain does not bring you back: \(0.5 \times 1.5 = 0.75\), a 25% loss. Exponential change works on the multiplier, every period, on whatever amount is there at the time.

Where it leads