Exponential Growth and Decay
Every question with a fully worked solution.
- Exponential Growth and Decay - Video - Exponential Growth and Decay Watch
Theory
In exponential growth and decay a quantity is multiplied by the same factor in every equal time period. This Year 10A maths page shows how to write and use growth and decay models, work with half-life and doubling time, compare linear and exponential change, find a growth rate with an \(n\)th root and find a time by trial and improvement.
A quantity shows exponential growth when it is multiplied by the same number, the growth factor, in every equal time period. Growing by \(r\%\) each period means multiplying by \(1 + \dfrac{r}{100}\): \(6\%\) growth has growth factor \(1.06\).
Exponential decay is the same idea with a factor between \(0\) and \(1\). Falling by \(8\%\) each period leaves \(92\%\), a factor of \(0.92\). The amount gets closer and closer to \(0\) but never reaches it.
The doubling time is the time it takes a growing quantity to double. The half-life is the time it takes a decaying quantity to halve. Both stay the same however much there is.
Linear change adds the same amount each period (constant differences); exponential change multiplies by the same factor (constant ratios). In a table, divide each value by the one before: if the answers are all equal, the relationship is exponential. Exponential growth always overtakes linear growth in the end.
For a starting amount \(A_0\) and \(t\) time periods:
\[A = A_0(1 + r)^t \qquad A = A_0(1 - r)^t\]for growth and decay at rate \(r\) (as a decimal) per period. With a doubling time \(d\) or a half-life \(h\):
\[A = A_0 \times 2^{t/d} \qquad A = A_0 \times \left(\dfrac{1}{2}\right)^{t/h}\]How to build and use a model
- Find the starting amount \(A_0\) (the value at \(t = 0\)).
- Find the factor: \(1 + r\) for growth, \(1 - r\) for decay, or use \(2^{t/d}\) or \(\left(\frac{1}{2}\right)^{t/h}\).
- Substitute the value of \(t\) and evaluate the power first, then multiply. Round only at the end, to a sensible unit (whole animals, cents).
Finding the time
- Count halvings or doublings when the numbers allow: \(960 \to 60\) is \(4\) halvings.
- Same base: divide by \(A_0\), write both sides as powers of the same base and equate the indices.
- Table of values: otherwise try whole numbers of periods until the amount first passes the target.
Working backwards and comparing
- Original amount: divide by the factor once for each period. Do not subtract the percentage.
- Two models: tabulate both and find the first period in which one passes the other.
(ii) A colony of \(5000\) seabirds shrinks by \(8\%\) each year. Write a rule and find the number after \(3\) years.
(i) Growth of \(6\%\): the factor is \(1 + 0.06 = 1.06\).
| \(P\) | \(=\) | \(800 \times 1.06^n\) |
| \(P(4)\) | \(=\) | \(800 \times 1.06^4\) |
| \(\) | \(=\) | \(1009.98\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{1010} \text{ cattle}\) |
(ii) A fall of \(8\%\) leaves \(92\%\): the factor is \(0.92\).
| \(N\) | \(=\) | \(5000 \times 0.92^n\) |
| \(N(3)\) | \(=\) | \(5000 \times 0.92^3\) |
| \(\) | \(=\) | \(3893.44\) |
| \(\) | \(\approx\) | \(\mathbf{3893} \text{ seabirds}\) |
(ii) A bacteria culture of \(150\) cells doubles every \(3\) hours. How many cells are there after \(12\) hours, and when will there be \(9600\)?
(i) \(20 \div 5 = 4\) half-lives, so halve \(4\) times:
| \(A\) | \(=\) | \(960 \times \left(\tfrac{1}{2}\right)^{20/5}\) |
| \(\) | \(=\) | \(960 \times \tfrac{1}{16} = \mathbf{60}~\textbf{mg}\) |
(ii) \(12 \div 3 = 4\) doublings. For the time, divide by \(150\) and write a power of \(2\):
| \(N\) | \(=\) | \(150 \times 2^{4} = \mathbf{2400}\) |
| \(150 \times 2^{t/3}\) | \(=\) | \(9600\) |
| \(2^{t/3}\) | \(=\) | \(64 = 2^6\) |
| \(t\) | \(=\) | \(\mathbf{18}~\textbf{hours}\) |
Plan A is linear: \(A = 500 + 40w\). Plan B is exponential: \(B = 500 \times 1.06^w\). Make a table near where they meet.
| \(w\) | \(9\) | \(10\) | \(11\) |
|---|---|---|---|
| A | \(860\) | \(900\) | \(940\) |
| B | \(844.74\) | \(895.42\) | \(949.15\) |
Plan B is first ahead after \(\mathbf{11}\) weeks. Plan A grows by the same \(\$40\) every week, but Plan B's weekly increase keeps growing, so it always catches up in the end.
(ii) A pond holds \(800\) L of water and loses \(10\%\) of it each day. Use trial and improvement to find, to \(1\) decimal place, when half the water is left.
(i) Substitute, divide, then take the square root (the \(n\)th root for \(n\) years):
| \(400\,000(1 + r)^2\) | \(=\) | \(484\,000\) |
| \((1 + r)^2\) | \(=\) | \(1.21\) |
| \(1 + r\) | \(=\) | \(\sqrt{1.21} = 1.1\) |
| \(r\) | \(=\) | \(\mathbf{10\%} \text{ p.a.}\) |
(ii) Half is left when \(0.9^t = 0.5\). Try values of \(t\):
| \(0.9^6\) | \(=\) | \(0.531\ldots \ (\text{too big})\) |
| \(0.9^7\) | \(=\) | \(0.478\ldots \ (\text{too small})\) |
| \(0.9^{6.5}\) | \(=\) | \(0.504\ldots\) |
| \(0.9^{6.6}\) | \(=\) | \(0.498\ldots\) |
So half the water is left after about \(\mathbf{6.6}\) days. This is the half-life of the water level.
Common pitfalls
Frequently asked questions
What is the difference between exponential growth and decay?
Both multiply by the same factor each period. In growth the factor is more than \(1\), so the amount increases faster and faster. In decay it is between \(0\) and \(1\), so the amount falls towards \(0\).
What is a half-life?
The time it takes for a decaying amount to halve. After two half-lives a quarter is left, after three an eighth, and so on.
How can you tell from a table whether growth is linear or exponential?
Subtract each value from the next: equal differences mean linear. Divide each value by the one before: equal ratios mean exponential.
How do you write a percentage increase as a growth factor?
Add the percentage to \(100\%\) and write it as a decimal: a \(3.5\%\) increase is a factor of \(1.035\). For a decrease, subtract it: a \(3.5\%\) decrease is \(0.965\).
Why does exponential growth overtake linear growth?
Linear growth adds the same amount each period. Exponential growth adds a fixed percentage of an amount that keeps getting bigger, so its increases keep getting bigger and eventually pass any fixed amount.