1. Various Forms of Exponential Functions
Doubling
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
Half-life
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

2. A = P(1 + i)n 1 Compound Interest
A is the final amount including interest and principal
A = P(1 + i)n
P is the original principal invested
i is the interest rate per compounding period
n is the number of compounding periods 1

3. A(t) = c(a)x Exponential Function (General Form)
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

4. Applications of Exponential Functions
Example 1
Two people are sent an at 12 p.m.
At 1:00 p.m. they send the to 3 people.
At 2:00 p.m. they send the to each of 3 people.
At 3:00 p.m. they send the to each of 3 people. etc, etc.
hour 1 2 3 4 5 6
number of people 2 6 18 54
162
486
1458

5. Equivalent expression
hour 1 2 3 4 5 6
number of people 2 6 18 54
162
486
1458
Hour
People told
Equivalent expression 1 2 3 4 5 6 2 2
2×3 6
2×32 18 54
2×33
162
2×34
486
2×35
1458
2×36
N = 2×3t

6. A(t) = c(a)t , a > 1 Exponential Growth N = c(a)t
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)
= bacteria
= 50(32)
= 1600 bacteria

7. A(t) = c(a)t = 25(2)5 = 25(32) = 800 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

8. N = c(a)t 2 = 1(a)2 2 = a2 = 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 after 2 hours.
N = c(a)t
2 = 1(a)2
2 = a2
= 800 bacteria

9. Determining the Growth Rate
Example 4: In the year 2000, the population of a town was people. In 2002, the population was 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
A(t) = c(a)t
28090 = c(1.06)2
28090 = c(a)2
= a2
31562 = c(a)4
(divide to find a)
25000 = c
1.06 = a

10. The truck depreciates in value by 20% per year.
Exponential Decay
A(t) = c(a)t , a < 1
c = initial amount a = decay rate
Example 5: After 5 years, a $ truck is worth $ What is the decay rate?
A(t) = c(a)t
= 64000(a)5
0.80 = a
The truck depreciates in value by 20% per year.
= a5

11. Ex. 6 The half-life of a substance is 4 hours
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)
(ii)
M = grams
M = 8 grams

12. Ex 7: A bacteria dish begins with 100 bacteria
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)
the doubling period is 1.25 hours.

13. Ex 7(con’d): A bacteria dish begins with 100 bacteria
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.
(iii)
(ii)
M = 100(2)6.4
= 100(84.44)
= 8444
After 8 hours there will be 8444 bacteria.