In a model N = N₀ekt, the number k is the growth constant: a rate of continuous change. It is not the same as the percentage increase per time period, which is (ek − 1) × 100%.
This lesson sits between the log laws of combining logarithms and the practical questions in the mixed practice set. It pulls together the ideas in exponential and logarithmic reasoning.
How are the two forms connected?
A quantity growing by p% each time period is multiplied by the factor (1 + p/100) each period. After t periods:
N = N₀ × (1 + p/100)t
Writing 1 + p/100 as ek gives N = N₀ekt. So the two constants are linked by:
| Direction | Formula |
|---|---|
| From percentage to k | k = ln (1 + p/100) |
| From k to percentage | p = (ek − 1) × 100 |
The two forms describe the same curve. The difference is only in which number you read off: k describes continuous change, p describes the jump over one full period.
Worked example
A population is modelled by N = 500e0.04t, where t is in years.
(a) Percentage growth per year. The yearly factor is e0.04 = 1.0408. So growth is about 4.08% per year, not 4%.
(b) Population after 10 years. N = 500e0.4 = 500 × 1.4918 = 745.9, so about 746.
(c) Doubling time. Solve 500e0.04t = 1000, so e0.04t = 2. Take ln: 0.04t = ln 2, so t = 0.6931 ÷ 0.04 = 17.33. The population doubles after about 17.3 years.
Check against the wrong reading. If 4% a year were used instead, 500 × 1.0410 = 740.1, which is about 6 less than 746. So the two readings do differ, and only the e form matches the stated model.
The mistake to watch for
The usual slip is to read k as the percentage directly.
Mistaken working: “k = 0.04, so the population grows by 4% each year.”
The student equated a continuous rate with a yearly percentage.
The correction is to find the yearly factor first. After one year, N = 500e0.04 = 520.4, which is an increase of 20.4 on 500, so 4.08%. For small k the difference is slight, but for larger k it grows: k = 0.5 gives e0.5 = 1.649, which is a 64.9% rise each period, not 50%.
Check yourself
1. For N = 200e0.05t, find the percentage growth per year to 3 significant figures.
Show answer
e0.05 = 1.0513, so the yearly growth is 5.13%.
2. A value falls by 8% each year, so N = N₀(0.92)t. Write this as N = N₀ekt and give k to 3 significant figures.
Show answer
Set ek = 0.92, so k = ln 0.92 = −0.0834. The negative sign shows decay.
3. An investment grows by 2% each year. After how many years does it first double? Give your answer to 3 significant figures.
Show answer
Solve 1.02t = 2, so t = ln 2 ÷ ln 1.02 = 0.6931 ÷ 0.01980 = 35.0. The investment doubles after 35.0 years.
Where this leads next
Take the ideas through the mixed practice set, then look back at linearising an exponential relationship to see how the same constants appear on a log graph.
If you can do each calculation but are unsure which form the question wants, a teacher can look at how you read the wording in online one-to-one Additional Mathematics tuition.