1
One model covers growth and decay.
A = A₀ · bt/T — start, factor, and how long one factor takes.
A₀
the starting amount
the value at t = 0
b
2 to double, 12 to halve
3 to triple, and so on
tT
how many periods have passed
T is one period
Getting tT right is nearly the whole job. The exponent counts periods, and
the units of t and T must match before you divide.
2
Half-life is the same model with b = 12.
A = 80(12)t/5
start 80, halves every 5 years
t = 15, so t5 = 3
three half-lives
A = 80(18) = 10 g
80 → 40 → 20 → 10
Three half-lives is not "all of it"
Halving three times removes seven eighths, not the lot. Nothing with a half-life
ever reaches zero, which is the same asymptote fact as in the graphing lesson,
wearing different clothes.
3
A percentage rate becomes the base directly.
grows 8% a year
A = A₀(1.08)t
keeps 100% and adds 8
falls 20% a year
A = A₀(0.80)t
keeps 80%
falls 5% a month
A = A₀(0.95)t
t counts MONTHS
Here the period is one, so tT is just t. Mixing a monthly rate with t in years is the
error that produces answers off by a factor of a thousand.
4
Substitute, then read the answer in context.
P = 20000(1.03)t
3% growth
= 20000(1.3439) ≈ 26,878
people, so round sensibly
Not 26,000: growing 3% ten times over is more than growing 30% once, because each
year's 3% is taken from a larger number. That gap is the whole point of compounding.
5
Variations: the half-life asked for, a part of a period, the year it halves.
The half-life is the unknown: 72 g falls to 9 g in 15 days.
72 → 36 → 18 → 9
count the halvings
3 half-lives in 15 days
972 = 18 = (12)³
half-life = 15 ÷ 3 = 5 days
time per period
Run it backwards by counting periods first, then dividing the time by that count.
Part of a period: 100 bacteria double every 2 hours. How many after 5 hours?
tT = 52 = 2.5
two and a half doublings
A = 100 · 22.5
the exponent need not be whole
≈ 100 × 5.657 ≈ 566
22.5 = 4√2
Between 400 (after 4 hours) and 800 (after 6 hours), as it must be. Halfway in time is not halfway in number: 566, not 600.
The year it first halves: a ₱40,000 gadget loses 20% a year.
keeps 80%, so A = 40000(0.8)t
the base is 0.80
0.8³ = 0.512, so after 3 years ₱20,480
still more than half
0.8⁴ = 0.4096, so after 4 years ₱16,384
now below ₱20,000
It first drops below half during the 4th year — not after 2½ years, which is what "20% twice and a half" would suggest.
6
When the unknown is the exponent, take logs.
20000 = 10000(1.06)t
set up the equation
2 = (1.06)t
divide by the start
log 2 = t · log 1.06
power law brings t down
t = 0.30100.0253 ≈ 11.9 years
so 12 years to be safe
4,000 bacteria after 9 hours
The power law is what makes this possible: bringing the exponent down in front turns
an unsolvable exponential into a one-line division. That single move is why
logarithms were invented.
On the exam
Periodical exams ask for an amount after a whole number of doublings or half-lives, or for the base of a percent model; entrance tests use periods longer than one unit (doubles every 4 hours) and percent drops applied year on year. Name A₀, b and T, then count periods as t ÷ T. For a 10-second check, the answer must sit between the values for the whole periods on either side.
Remember the flow
Name A₀, b and T · Exponent is tT · Match the units · Unknown exponent means take logs
A percentage rate is already the base: 8% growth is 1.08, 20% loss is 0.80.
