# EXPONENTIAL GROWTH AND DECAY WORD PROBLEMS

Problem 1 :

The weight of bacteria in a culture t hours after it has been established is given by

W= 2.5 × 20.04t

grams. After what time will the weight reach

a. 4 grams     b. 15 grams

Solution:

a) Weight = 4 grams

4 = 2.5 × 20.04t

4/2.5 = 20.04t

1.6 = 20.04t

log 1.6 = 0.04t log 2

0.04t = log 1.6/log 2

0.04t = 0.2041/0.3010

0.04t = 0.6780

t = 0.6780/0.04

t = 16.95

Approximately 17 hours.

b)  Weight = 15 grams

15 = 2.5 × 20.04t

15/2.5 = 20.04t

6 = 20.04t

log 6 = 0.04t log 2

0.04t = log 6/log 2

0.04t = 0.7781/0.3010

0.04t = 2.585

t = 2.585/0.04

t = 64.62

Approximately 64.6 hours.

Problem 2 :

The population of bees in hive, t hours after it has been discovered, is given by

Pn = 5000 x 20.09t

After what time the population will reach

a. 15000    b. 50000

Solution :

a)  Pn = 5000 x 20.09t

15000 = 5000 x 20.09t

20.09t 15000/5000

20.09t = 3

log (20.09t) = log 3

0.09t log 2 = log 3

0.09t = log 3 / log 2

0.09t = 0.4771/0.3010

0.09t = 1.58

t = 17.55

Approximately 17.6 hours.

b)  Pn = 5000 x 20.09t

50000 = 5000 x 20.09t

20.09t = 50000/5000

20.09t = 10

log (20.09t) = log 10

0.09t log 2 = 1

0.09t = 1/0.3010

t = 36.9 hours

Problem 3 :

The weight Wt grams of radioactive substance remaining after t years is given by

Wt = 1000 x 2-0.03t grams. Find

i)  The initial weight

ii)  The weight after

a) 10 years    ii) 100 years     iii) 1000 years

iii)  Graph Wt against t using a and b only

Solution :

Wt = 1000 x 2-0.03t

i) Initial weight, when t = 0

W0 = 1000 x 2-0.03(0)

W0 = 1000 grams

ii)  The weight after

a) 10 years    ii) 100 years     iii) 1000 years

 when t = 10W10 = 1000 x 2-0.03(10)W10 = 1000 x 2-0.3W10 = 1000 x 0.8130W10 = 813 grams when t = 100W10 = 1000 x 2-0.03(100)W100 = 1000 x 2-3W100 = 1000 x 0.125W100 = 125 grams

when t = 1000

W1000 = 1000 x 2-0.03(1000)

W1000 = 1000 x 2-30

W1000 = 1000 x 9.313 x 10-10

W1000 = 103 x 9.313 x 10-10

W1000 = 9.313 x 10-7

Problem 4 :

The weight of radioactive substance after t years is given by the formula

Wt = W0  x 3-0.003t grams. Find

i)  The initial weight

ii)  The percentage remaining after

a) 100 years    ii) 500 years     iii) 1000 years

Solution :

i) Initial weight, when t = 0

Wt = W0  x 3-0.003(0)

Wt = Wgrams

ii)  The weight after

i) 100 years    ii) 500 years     iii) 1000 years

i)  when t = 100

W10 = W0 x 3-0.003(100)

W10 = W0 x 3-0.3

W10 = W0 x 0.719

Converting into percentage, we get

71.9 % of W0

ii)  when t = 500

W500 = W0 x 3-0.003(500)

W500 = W0 x 3-1.5

W500 = W0 x 0.192

Converting into percentage, we get

19.2 % of W0

iii)  when t = 1000

W1000 = W0 x 3-0.003(1000)

W1000 = W0 x 3-3

W1000 = W0 x 0.03703

Converting into percentage, we get

3.7 % of W0

Problem 5 :

The temperature Tt (°C) of a liquid which has been placed in a refrigerator is give by

T= 100 x 2-0.04t

a) the initial temperature

b)  The temperature after

i) 10 minutes     ii)  25 minutes      iii)  60 minutes

c) Find how long it takes for the temperature to reach 20°C.

Solution :

T= 100 x 2-0.04t

a) Initial temperature, when t = 0

T= 100 x 2-0.04(0)

T= 100

b)  The temperature after

i) 10 minutes

T10 = 100 x 2-0.04(10)

T10 = 100 x 2-0.4

T10 = 75.78

ii)  25 minutes

T25 = 100 x 2-0.04(25)

T25 = 100 x 2-1

T25 = 50

iii)  60 minutes

T60 = 100 x 2-0.04(60)

T60 = 100 x 2-2.4

T60 = 18.94

Problem 6 :

The weight Wt (°C) of a radioactive substance remaining after t years is given by the formula

W= W0 x 2-0.0005t

a) the original weight

b)  The weight loss after 1200 years

Solution :

a)  W= W0 x 2-0.0005t

Original weight, when t = 0

W= W0 x 2-0.0005(0)

Initial weight is W0

b)  The weight loss after 1200 years

When t = 1200

W1200 = W0 x 2-0.0005(1200)

W1200 = W0 x 0.9597

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