P(t) = P₀·rt · geometric sequences · doubling · changing the time base
Fixes these misses: M1 Q14 (doubling), M1 Q19 (geometric sequence), M2 Q10 (rate per period).
Core concepts
The exponential model
P(t) = P₀ · rt
P₀ is the starting amount; r is the growth factor per one unit of t.
Growth of p% per period → r = 1 + p/100. Decay of p% → r = 1 − p/100.
r = 2 means doubling each period; r = ½ means halving.
Doubling / halving every k units
P(t) = P₀ · 2(t/k)
If it doubles every k time units, the exponent is t/k, not t.
Example: doubles every 3 hours, after 15 hours → 2(15/3) = 25 = 32×.
Geometric sequences
aₙ = a₁ · r(n−1)
"Each term is R times the one before" → geometric with ratio r = R.
The exponent is (n − 1) because the first term uses r0 = 1.
Changing the time base
To convert a per-year rate to per-18-months, raise the yearly factor to the 1.5 power.
Per-period factor = r(period length in years); then percent change = (factor − 1)×100.
Worked example — rate per period
Problem. P(t) = 290·(1.0446)t models population t years after 2005. It increases by n% every 18 months. Find n.
18 months = 1.5 years, so the growth factor over one period is (1.0446)1.5.
(1.0446)1.5 ≈ 1.0676.
That’s about a 6.76% increase → n ≈ 6.8 (this is the M2 Q10 answer).
Watch out: Don’t read r directly as the answer. r = 1.0446 is the yearly factor; the question asks per 18 months, so you must raise it to the 1.5 power first. Reading "4.46%" straight off is the trap that cost the point.
Video lessons
Curated walkthroughs from trusted math channels — watch one or two before doing the worksheet.