Download Exponential Functions

Various Forms of Exponential
Functions
Doubling
M  c2
Half-life
t
d
1
M  c 
2
M is the total number after time t.
c is the initial amount at time t = 0.
t is the time
d is the doubling period
t
h
M is the total number after time t.
c is the initial amount at time t = 0.
t is the time
h is the half-life
Compound Interest
A = P(1 + i)n
1
A is the final amount including
interest and principal
P is the original principal invested
i is the interest rate per
compounding period
n is the number of
compounding periods
Exponential Function (General Form)
A(t) = c(a)x
A is the total amount or number
(at time t)
c is the initial amount or number
a is the growth factor or decay rate
x is the number of growth or
decay periods
Applications of Exponential Functions
Example 1
Two people are sent an email at 12 p.m.
At 1:00 p.m. they send the email to 3 people.
At 2:00 p.m. they send the email to each of 3 people.
At 3:00 p.m. they send the email to each of 3 people. etc, etc.
hour
0
1
2
3
4
5
6
number
of people
2
6
18
54 162 486 1458
hour
0
1
2
3
number
of people
2
6
18
54 162 486 1458
Hour
0
1
2
3
4
5
6
People told
2
6
18
54
162
486
1458
4
5
Equivalent
expression
2
2×3
2×32
2×33
2×34
2×35
2×36
N = 2×3t
6
Exponential Growth A(t) = c(a)t , a > 1
c = initial amount
a = growth rate
Example 2: A bacteria dish begins with 50 bacteria. If
the doubling period is 1 hour, how many bacteria are
there after (i) 5 hours? (ii) 10 hours?
N = c(a)t
(i) N = 50(2)t
(ii) N = 50(2)10
= 50(2)5
= 50(1024)
= 51 200 bacteria
= 50(32)
= 1600 bacteria
Example 3: A bacteria dish begins with 25 bacteria. If
the doubling period is 2 hours, how many bacteria are
there after 10 hours?
Solution 1: There are five doubling periods in 10 hours
A(t) = c(a)t
= 25(2)5
= 25(32)
= 800 bacteria
Example (continued): A bacteria dish begins with 25
bacteria. If the doubling period is 2 hours, how many
bacteria are there after 10 hours?
Solution 2: If you start with one bacteria, there are
2 after 2 hours.
N = c(a)t
2 = 1(a)2
2 = a2
2 a
N  25( 2)
10
1
2 10
 25(2 )
 25(32)
= 800 bacteria
Determining the Growth Rate
Example 4: In the year 2000, the population of a
town was 28 090 people. In 2002, the population was
31 562. What is the growth rate? What was the
population in 1998?
Let 1998 be the initial time period when t = 0.
A(t) =
c(a)t
31562 c(a) 4

28090 c(a)2
28090 = c(a)2
31562 = c(a)4
1.1236 = a2
(divide to find a)
1.1236  a
1.06 = a
A(t)= c(a)t
28090 = c(1.06)2
28090
c
2
1.06
25000 = c
Exponential Decay A(t) = c(a)t , a < 1
c = initial amount
a = decay rate
Example 5: After 5 years, a $64 000 truck is
worth $20 971.52. What is the decay rate?
A(t) = c(a)t
5
0.32768  a
20971.52 = 64000(a)5
0.80 = a
20971.52
 a5
64000
0.32768 = a5
The truck depreciates in
value by 20% per year.
Ex.6 The half-life of a substance is 4 hours.
How much of the substance will remain after
(i) 12 hours (ii) 2 days if there were 64 grams at
the start?
(i)
1
M  c 
2
t
h
1

(ii) M  64   
2
12
4
1
M  64   
2
1
M  64   
8
M = 8 grams
48
4
12
1
M  64   
2
 1 
M  64  

 4096 
M = 0.0156 grams
Ex 7: A bacteria dish begins with 100 bacteria.
After 5 hours there are 1600 bacteria.
(i) What is the doubling period?
(ii) Find the population after t hours.
(iii) Determine the number of bacteria after 8 h.
(i) M  c  2
t
d
1600  100  2
16  2
5
d
2 2
4
5
d
5
d
5
4
d
4d  5
d  1.25
the doubling period
is 1.25 hours.
Ex 7(con’d): A bacteria dish begins with 100
bacteria. After 5 hours there is 1600 bacteria.
(i) What is the doubling period?
(ii) Find the population after t hours.
(iii) Determine the number of bacteria after 8 h.
(ii) M  c  2
8
1.25
t
d
(iii) M  100  2
t
1.25
M  100  2
M = 100(2)6.4
= 100(84.44)
= 8444
After 8 hours there will
be 8444 bacteria.