1. The document is a math assignment submitted by Amos Kiprotich Meli for MATH 101 Precalculus at the University of Eastern Africa, Baraton. 2. The assignment contains 3 problems involving modeling population growth and decline using exponential functions. The first problem looks at bacterial growth over time, the second models population decline due to economic factors, and the third uses an exponential growth model to determine population size at various time points and growth rates. 3. The assignment is solved showing the steps and solutions for each part of the 3 problems involving exponential modeling of real-world population changes over time.

1. UNIVERSITY OF EASTERN AFRICA, BARATON
SCHOOL OF SCIENCE AND TECHNOLOGY.
DEPARTMENTOF MATHEMATICS, CHEMISTRY AND PHYSICS.
An assignmentwritten in partial fulfillment of the course;
MATH 101 PRECALCULUS.
TOPIC: MATH 101 ASSIGNMENT
NAME:AMOS KIPROTICH MELI
ID: SAMOME2111
INSTRUCTOR: PROF PAUL FRANCIS
ASSSIGNMENT.
1) Bacteria are growing in a culture, and their number is increasing at the rate of
8% an hour.
Initially 800 bacteria are present.
a) Determine an equation that gives the numberN, of the bacteria present after t
hours.
Solution
P = S (1 + r) n
N = 800 (1 + 8/100) t
N= 800(1.08) t
b) How many bacteria are present after one hour? (Give your answer to the
nearest integer) Solution
P = S (1 + r) n

2. = 800 (1 + 8/100) 1
Ans. 864 bacteria
c) How many bacteria are present after 7 hours? (Give your answer to the nearest
Integer)
Solution
P = S (1 + r) n
= 800 (1 + 8/100) 7
Ans. 1371 bacteria
2) Because of an economic downturn, the population of a certain urban area
declines at the rate of 2% per year. Initially, the population is 650,000. To the
nearest person, what is the population after 4 years?
Solution
P = S (1 - r) n
= 650,000 (1-2/100)4
Ans. 599,539 people
3) At the start of an experiment 2000 bacteria are present in a colony. Two hours
Later the population is 3800. N=N0e kt(10mks)
Where;
N = Population at time t
N0= the size of population at time t = 0
k = Positive constant.
a) Determine the growth constant k
Solution N=N0ekt
3800 = 2000ek (2)
3800/2000 = 2000/2000e2k loge1.9 =
2k logee
loge1.9= 2k (1)
2 2
k.= 0.3209
b) Determine the population 5 hours after the start of the experiment.
Solution
N=N0ekt
= 2000e0.3209(5)Ans.
9,950 bacteria
c) When will the population reach 10,000?
Solution
N=N0ekt

3. 10000 = 2000e0.3209(t) 10000/2000
= 2000/2000e0.3209(t)loge5 =
0.3209(t) logee loge5 = 0.3209(t)
×1 loge5/0.3209 = 0.3209(t)
/0.3209 t is approx. 5hours