Download L6- Problem Solving with Exponential Growth and Decay

Doubling Time and Half time
Doubling Time
 The time it takes for something to double in size
t
d where:
 Equation:
0
 P – Population
P  P (2)
 P0-Initial Population
 t- Time
 d – Doubling time
 2 – base for doubling
Example 1
 A bacterial culture began with 10 bacteria.
Its growth can bet modeled using the
formula B  10(2) 8 , where B is the number of
bacteria after t hours.
a) What is the doubling time?
b) How many bacteria are present after 8
hours?
c) How many bacteria are present after 16
hours?
Example 1
B  10(2)
t
8
a) What is the doubling time?
8 hours
b) How many bacteria are
present after 8 hours?
8
1
8
B

10
(
2
)

10
(
2
)
 10(2)  20
Substitute t=8
There are 20 bacteria after 8 hours
c) How many bacteria are present after 16 hours?
How does it relate to doubling?
16
8
2
B

10
(
2
)

10
(
2
)
 10(4)  40
Substitute t=16
There are 40 bacteria after 16 hours
It relates to doubling because the bacteria
double twice from the initial population
Half Life
 The time it takes for a quantity to decay or be reduced
to half its original amount.
1
M  M0 
 M – Final quantity
2
 Equation:
 M0-Initial quantity
 t- Time
 h – Half life
 1/2 – base for halving
t
h
where:
Example 2
 The half-life of Iodine-131 (Radioactive), is 8 days
where, with an initial amount of 50g, the final
t
quantity can be modeled by
 1 8
M  50 
2
 How much Iodine is left after:
 8 days?
 12 days?
Example 2 Solutions
 How much Iodine is left after:
 8 days?
 Substitute t=8
t
8
8
8
1
1
1
1
1
M  50   50   50   50   25
2
2
2
2
 There’s 25g of Iodine after 8 days.
 12 days?
12
8
1.5
1
1
M  50   50 
2
2
 500.354  17.68
 There’s 17.68g of Iodine after 12 days