Interest, Growth and Decay
Number
The model
- A quantity changing by a fixed percentage each period grows or decays exponentially
- The same relationship covers both directions: B = A × kⁿ, where A is the starting amount, B the amount after n periods, and k the multiplier
- A value of k above 1 gives growth, and a value between 0 and 1 gives decay
- k is never negative, since a multiplier below zero would flip the sign of the quantity
- The period need not be a year: questions can also use other periods, such as days
Where it appears
| Growth | Decay |
|---|---|
| Population size | Radioactive mass |
| Bacterial cultures | A cooling liquid |
| An investment compounding | An asset losing value |
Worked example
Exponential growth of a population
A colony of 500 bacteria grows by 8% every hour. Find the population after 10 hours.
Solution:
- Growth of a fixed percentage each period is exponential, so the same multiplier is applied once per period
- Growing by 8% leaves 108% of the population, so the multiplier is 1.08
- The rate is per hour and the time is in hours, so the power is 10: always match the power to the period the rate is quoted for
- Population = 500 × 1.08¹⁰
- 1.08¹⁰ = 2.158925…, so the population has slightly more than doubled
- 500 × 2.158925… = 1079.46
- Bacteria are counted in whole organisms, so the answer is about 1079
- The context fixes the form of the answer here: 1079.46 bacteria is not a possible population, so give a whole number
Rearranging the model
- Where the starting amount is unknown, divide the final amount by kⁿ
- Where the multiplier is unknown, divide the final amount by the starting amount and take the nth root
- Where the number of periods is unknown, test whole-number values until the target is first passed, since n counts completed periods
Worked example
Finding the yearly rate of compound interest
£5000 is invested at r% per year compound interest, and is worth £6100 after 7 years. Find r.
Solution:
- Let k be the yearly multiplier, so 5000 × k⁷ = 6100
- Divide both sides by 5000: k⁷ = 6100 ÷ 5000 = 1.22
- Take the 7th root: k = ⁷√1.22 = 1.028814…
- The multiplier is 1 + r/100, so r/100 = 0.028814…
- r = 2.88 to 3 significant figures
- Check: 5000 × 1.0288⁷ = 6099.39…, which is £6100 to the nearest pound
- Writing 1.0288 as the answer would give the multiplier, not the rate
Exam tip
Round a decimal number of periods up
If your working gives a decimal number of periods, such as n = 10.48, the answer is the next whole number, 11, even though 10.48 would normally round to 10: the target is first passed during the eleventh period. The schemes gave a decimal n every mark except the last, so finish with the whole number the question asks for.
Worked example
Whole hours until a quantity falls below a level
The mass of a drug in a patient's blood is 400 mg, and it falls by 18% each hour. After how many complete hours is the mass first below 50 mg?
Solution:
- Losing 18% leaves 82%, so the hourly multiplier is 0.82
- After n hours the mass is 400 × 0.82ⁿ mg
- A first trial: 400 × 0.82⁵ = 148.29… mg, still well above 50, so try larger values of n
- n = 10: 400 × 0.82¹⁰ = 54.98… mg, still above 50
- n = 11: 400 × 0.82¹¹ = 45.08… mg, now below 50
- 10 and 11 are the neighbouring values either side of 50 mg, so the mass is first below 50 mg after 11 hours
- Rounding down to 10 is wrong: after 10 hours the mass is still 54.98 mg