SME
Human Geography · Problem type
Population growth and doubling time
Use this for “A city of 1,000,000 grows 2 % a year. How many people will it have in 20 years, and when will it double?”
Chart
P(20) ≈ 1,486,000
This is an example. Type your own values to start a new problem.
Pₜ
=
P₀
×
(1
+
÷
100)
T₂
Doubling time
calculated · years
T₇₀
Rule-of-70 estimate
calculated · years
Formulas
Pₜ
=
P₀
×
(1 +
r
100
)
t
1,485,947
=
1,000,000
×
(1 +
2
100
)
20
T₂
=
ln(2)
ln(1 +
r
÷ 100)
35.0028
=
ln(2)
ln(1 + 2 ÷ 100)
T₇₀
=
70
r
35
=
70
2
Assumptions
•
The growth rate r stays the same every year, and each year multiplies the population by 1 + r ÷ 100.
•
r is the yearly change from births, deaths and migration together, as a percent; a negative r is a decline.
•
Rule of 70: ln 2 is about 0.693, so for small r the doubling time is about 70 ÷ r years.
•
A population that is not growing (r ≤ 0) never doubles, so it has no doubling time.
Step-by-step
WE KNOW
P₀ = 1,000,000, r = 2%, t = 20 years
FIND
projected population (Pₜ), doubling time (T₂), rule-of-70 estimate (T₇₀)
Step 1 · Find projected population (Pₜ)
Use Pₜ = P₀ × (1 + r ÷ 100)ᵗ
Each
year
multiplies
the
population
by
1
+
r
100,
so
t
years
multiply
it
by
that
factor
t
times.
Pₜ
=
1,000,000
×
(1 +
2
100
)
20
1
+
2
100
=
1.02
1.02
20
=
1.485947
1,000,000 × 1.485947 = 1,485,947
Pₜ
=
1,485,947
Step 2 · Find doubling time (T₂)
Use T₂ = ln(2) ÷ ln(1 + r ÷ 100)
Doubling
means
the
growth
factor
reaches
2:
solve
(1 +
r
100
)
T
₂
=
2
by
taking
ln
of
both
sides.
T₂
=
ln(2)
ln(1 + 2 ÷ 100)
ln(2) = 0.6931
ln(1
+
2
100
)
=
0.0198
T₂
=
35.0028
years
Step 3 · Find rule-of-70 estimate (T₇₀)
Use T₇₀ = 70 ÷ r
Rule
of
70:
100
×
ln
2
is
about
70,
and
ln(1
+
r
100
)
is
about
r
100
when
r
is
small.
T₇₀
=
70
2
T₇₀
=
35
years
Check
Put the numbers back in. Both sides should match.
1,485,947
=
1,000,000
×
(1 +
2
100
)
20
✓
35.0028
=
ln(2)
ln(1 + 2 ÷ 100)
✓
35
=
70
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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