MATHEMATICS
IN THE
MODERN
WORLD
PATTERNS IN NATURE
AND REGULARITIES IN
THE WORLD
Solving Word Problems Involving
Exponential Growth in Population
The exponential growth model A = 30e0.02t
describes the population of a city in the
Philippines in thousands, t years after 1995.
a. What was the population of the city in
1995?
b. What will be the population of the city in
2017?
SOLUTION:
a. Since our exponential growth models describe the
population t years after 1995, we consider 1995 as t = 0 and
then solve for A, our population size.
A = 30e0.02t
A = 30e(0.02)(0)
Replace t with t = 0
A = 30e0
A = 30(1)
e0 = 1
A = 30
Therefore, the city population in 1995 was 30, 000.
SOLUTION:
b. We need to find A for the year 2017. To find t, we subtract 2017
and 1995 to get t = 22, which we then plug in our exponential
growth model.
A = 30e0.02t
A = 30e(0.02)(22)
Replace t with t = 22
A = 30e0.44
A = 30(1.55271)
e0.44 ≈ 1.55271
A ≈ 46.5813
Therefore, the city population would be about 46, 581 in 2017.
using 2.718 ≈ 46, 579
c. What will be the population in 2020?
Solution:
A = 30e0.02t
A = 30e(0.02)(25)
A = 30e0.5
A = 30(1.64872)
A ≈ 49.4616
Replace t with t = 25
e0.5 ≈ 1.64872
Therefore, the city population would be about 49, 461 in
2020.
using 2.718 ≈ 49, 459
The exponential growth model A =
50e0.07t describes the population of a city in
the Philippines in thousands, t years after
1997.
a. What is the population after 20 years?
b. What is the population in 2037?
SOLUTION:
a. What is the population after 20 years?
A = 50e0.07t
A = 50e(0.07)(20)
Replace t with t = 20
A = 50e1.4
A = 50(4.05520)
e1.4 ≈ 4.05520
A ≈ 202.76
Therefore, the city population would be about 202, 760
after 20 years.
using 2.718 ≈ 202, 731
SOLUTION:
a. What is the population in 2037?
A = 50e0.07t
A = 50e(0.07)(40)
Replace t with t = 40
A = 50e2.8
A = 50(16.44465)
e2.8 ≈ 16.44465
A ≈ 822.2325
Therefore, the city population would be about 822, 233
in 2037.
using 2.718 ≈ 821, 993
EXERCISES:
Determine what comes next in the given
patterns.
1. A, C, E, G, I, K
2. 15 10 14 10 13 10 12
3. 3 6 12 24 48 96 192
4. 27 30 33 36 39 42
5. 41 39 37 35 33 31
Substitute the given values in the formula A = Pert
to find the missing quantity.
6. P = 680, 000; r = 12% per year; t = 8 years
7. A = 1, 240, 000; r = 8% per year; t = 30 years
8. A = 786, 000; P = 247, 000; t = 17 years
9. A = 731, 093; P = 525, 600; r = 3% per year.
6. P = 680, 000; r = 12% per year; t = 8 years
SOLUTION:
A=?
A = Pert
A = 680,000 e(0.12)(8)
A = 680,000 e0.96
A = 680,000 (2.61170)
e0.96 ≈ 2.61170
A ≈ 1, 775, 956
using 2.718 ≈ 1,775, 777
using scientific calculator ≈ 1,775, 954
7. A = 1, 240, 000; r = 8% per year; t = 30 years
SOLUTION:
P=?
A = Pert
1, 240, 000 = Pe(0.08)(30)
1, 240, 000 = Pe2.4
1, 240, 000 = P(11.02318)
e2.4 ≈ 11.02318
112, 490 ≈ P
using 2.718 ≈ 112, 518
using scientific calculator ≈ 112, 490
8. A = 786, 000; P = 247, 000; t = 17 years
SOLUTION:
A = Pert
786, 000 = 247, 000 e(r)(17)
786, 000 = 247, 000 e17r
786,000
247,000
= e17r
3.18219 = e17r
ln 3.18219 = 17r
ln 3.18219
17
= r
0.06809 ≈ r
Therefore the rate is approximately 6.81%
9. A = 731, 093; P = 525, 600; r = 3% per year.
SOLUTION:
A = Pert
731, 093 = 525, 600 e(0.03)(t)
731, 093 = 525, 600 e0.03t
731,093
525,600
= e0.03t
1.39097 = e0.03t
ln 1.39097 = 0.03t
ln 1.39097
0.03
=t
11. 00004 ≈ t
Therefore t is approximately 11 years.
Answer completely.
10. Suppose the population of a certain
bacteria in a laboratory sample is 100. And if it
doubles in population every 6 hours, what is
the growth rate? How many bacteria will there
be in two days?
Suppose the population of a certain bacteria in a laboratory
sample is 100. And if it doubles in population every 6 hours,
what is the growth rate? How many bacteria will there be in
two days?
To compute for the growth rate:
A = Pert
200 = 100 e(r)(6)
200 = 100 e6r
200
100
= e6r
2 = e6r
ln 2 = 6r
ln 2
6
=r
0.1155245301 ≈ r
Therefore the growth rate is approximately 11.55%
Suppose the population of a certain bacteria in a laboratory
sample is 100. And if it doubles in population every 6 hours,
what is the growth rate? How many bacteria will there be in
two days?
To compute for the number of bacteria in two days:
A = Pert
A = 100 e(0.1155245301)(48)
A ≈ 25, 600
There will be approximately 25, 600 bacteria in
two days.
using 100 x 28 = 25, 600
using 2.718 ≈ 25, 555
REFERENCE:
Mathematics in the Modern World Philippine Edition
Rex Bookstore