Exponential growth and decay

What is it about?

In exponential growth, a quantity changes by the same factor $b$ in equal time steps:

f(t) = a \cdot b^t \qquad b = 1 + \frac{p}{100}

$a$ is the initial value, $b$ the growth factor, $p$ the percentage change per time step.
For $b > 1$ , $f$ grows; for $0 < b < 1$ , $f$ decreases (exponential decay, e.g. radioactive decay).
For a decrease, $p$ is negative: a decrease of $8\,\%$ means $p = -8$ and $b = 1 - 0.08 = 0.92$ . So the growth factor is not $0.08$ .

In linear growth, on the other hand, the same amount is added in each time step. At first both look similar – in the long run, however, the exponential function grows much faster.

The doubling time (or half-life) is always the same in exponential growth. From $b^{T_d} = 2$ or $b^{T_{1/2}} = 0.5$ the logarithm gives:

T_d = \log_b 2 = \frac{\lg 2}{\lg b} \qquad T_{1/2} = \log_b 0.5 = \frac{\lg 0.5}{\lg b}

Instead of $\lg$ you can use any logarithm, e.g. $\ln$ . The simulation shows the value rounded to two decimal places (“≈”). Only when it is exact does it show “=” (e.g. $T_d = 1$ day for $p = 100$ ).

Try it

Start the water lilies with “Play”. The pond is full after 10 days. When was it half full?
Compare compound interest with linear growth (simple interest). 5 % interest
Watch the individual nuclei in radioactive decay. Do they decay evenly? Iodine-131 With $p = -8$ you get $T_{1/2} \approx 8.31$ days – close to the real value of just over 8 days.
Switch on the logarithmic y-axis. What happens to the graph? Logarithmic

Exercises

Exercise 1: The water lily puzzle

The water lilies double their area every day. After 10 days the pond is completely covered. After how many days was it half covered?

Show solution

After 9 days – on the last day the area doubles from half to full. Many people guess 5 days at first; that is exactly what makes exponential growth so hard to imagine. Show in the simulation

Exercise 2: Compound interest

$1000\,€$ is invested at $5\,\%$ interest, with the interest added every year. How much is in the account after 10 years? When has it doubled?

Show solution

$f(10) = 1000 \cdot 1.05^{10} \approx 1628.89\,€$ . Doubling time: $T_d = \frac{\ln 2}{\ln 1.05} \approx 14.2$ years. Without compound interest it would only be $1500\,€$ after 10 years. Show in the simulation

Exercise 3: Half-life

Iodine-131 has a half-life of about 8 days. What fraction is still there after 24 days? What is the daily decrease in percent?

Show solution

24 days are 3 half-lives: $\left(\frac{1}{2}\right)^3 = \frac{1}{8} = 12.5\,\%$ . Daily factor: $b = 0.5^{1/8} \approx 0.917$ , so a decrease of about $8.3\,\%$ per day.

Exercise 4: Rule of thumb

For small percentages there is a rule of thumb: doubling time $\approx \frac{70}{p}$ . Check it for $p = 2$ , $p = 5$ and $p = 10$ .

Show solution

Exact: $\frac{\ln 2}{\ln 1.02} \approx 35.0$ , $\frac{\ln 2}{\ln 1.05} \approx 14.2$ , $\frac{\ln 2}{\ln 1.1} \approx 7.3$ . The rule of thumb gives $35$ , $14$ and $7$ – very good for small $p$ .

Ideas for the classroom

Use: About grade 10 (exponential functions, exponential growth, logarithm).
Decay: Each of the 400 nuclei decays at a random time; on average the number follows the exponential function. This shows that decay is a random process.
Logarithmic axis: Exponential functions appear as straight lines – a good preparation for the logarithm.
Common pitfalls:
- Exponential growth is extrapolated like linear growth (water lily puzzle).
- “After two half-lives everything has decayed.” In fact a quarter is still left.
- Percentage and growth factor are mixed up: for an $8\,\%$ decrease, $b = 0.92$ , not $0.08$ or $1.08$ . The “Growth factor” readout shows both side by side.
Table of values: In the example “Free” a table of values replaces the picture. Have students compare the factors “· b” between the rows with the differences in linear growth.