1. EXPONENTIAL GROWTH
Exponential functions can be applied to real – world problems.
One instance where they are used is population growth.
The function for the population model is :
where: P = the number of individuals in the population at time t
A = the number of individuals in the population at time = 0
k = a positive constant of growth
e = the natural logarithm base
t = time in years

2. EXAMPLE :. A country’s population in 1994 was 107 million
EXAMPLE : A country’s population in 1994 was 107 million. In 1997 it was 112
million. Estimate the population in 2008 using the exponential
growth model. Round your answer to the nearest million.

3. EXAMPLE :. A country’s population in 1994 was 107 million
EXAMPLE : A country’s population in 1994 was 107 million. In 1997 it was 112
million. Estimate the population in 2008 using the exponential
growth model. Round your answer to the nearest million.
P = 112 ( population now )
A = 107 ( population then )
t1 = 3 years ( 1994 – 1997 )
t2 = 14 years ( 1994 – 2008 )

4. EXAMPLE :. A country’s population in 1994 was 107 million
EXAMPLE : A country’s population in 1994 was 107 million. In 1997 it was 112
million. Estimate the population in 2008 using the exponential
growth model. Round your answer to the nearest million.
P = 112 ( population now )
A = 107 ( population then )
t1 = 3 years ( 1994 – 1997 )
t2 = 14 years ( 1994 – 2008 )
** we first have to find “k” by substitution using t1 , A , and P

5. EXAMPLE :. A country’s population in 1994 was 107 million
EXAMPLE : A country’s population in 1994 was 107 million. In 1997 it was 112
million. Estimate the population in 2008 using the exponential
growth model. Round your answer to the nearest million.
P = 112 ( population now )
A = 107 ( population then )
t1 = 3 years ( 1994 – 1997 )
t2 = 14 years ( 1994 – 2008 )

6. EXAMPLE :. A country’s population in 1994 was 107 million
EXAMPLE : A country’s population in 1994 was 107 million. In 1997 it was 112
million. Estimate the population in 2008 using the exponential
growth model. Round your answer to the nearest million.
- divide both sides by 107

7. EXAMPLE :. A country’s population in 1994 was 107 million
EXAMPLE : A country’s population in 1994 was 107 million. In 1997 it was 112
million. Estimate the population in 2008 using the exponential
growth model. Round your answer to the nearest million.
- divide both sides by 107
- take “ln” by both sides
This gives us “k”…

8. EXAMPLE :. A country’s population in 1994 was 107 million
EXAMPLE : A country’s population in 1994 was 107 million. In 1997 it was 112
million. Estimate the population in 2008 using the exponential
growth model. Round your answer to the nearest million.
P = 112 ( population now )
A = 107 ( population then )
t1 = 3 years ( 1994 – 1997 )
t2 = 14 years ( 1994 – 2008 )
k = .015
** we can now find P by substitution using t2 , A , and k

9. EXAMPLE :. A country’s population in 1994 was 107 million
EXAMPLE : A country’s population in 1994 was 107 million. In 1997 it was 112
million. Estimate the population in 2008 using the exponential
growth model. Round your answer to the nearest million.
P = 112 ( population now )
A = 107 ( population then )
t1 = 3 years ( 1994 – 1997 )
t2 = 14 years ( 1994 – 2008 )
k = .015
P = 107 e14•0.015

10. EXAMPLE :. A country’s population in 1994 was 107 million
EXAMPLE : A country’s population in 1994 was 107 million. In 1997 it was 112
million. Estimate the population in 2008 using the exponential
growth model. Round your answer to the nearest million.
P = 107 e14•0.015
P = 107 e0.21
P = 107 • ( )
P =
Multiplied 14 times 0.015

11. EXAMPLE :. A country’s population in 1994 was 107 million
EXAMPLE : A country’s population in 1994 was 107 million. In 1997 it was 112
million. Estimate the population in 2008 using the exponential
growth model. Round your answer to the nearest million.
P = 107 e14•0.015
P = 107 e0.21
P = 107 • ( )
P =
- Evaluated e0.21

12. EXAMPLE :. A country’s population in 1994 was 107 million
EXAMPLE : A country’s population in 1994 was 107 million. In 1997 it was 112
million. Estimate the population in 2008 using the exponential
growth model. Round your answer to the nearest million.
P = 107 e14•0.015
P = 107 e0.21
P = 107 • ( )
P =
- multiplied

13. EXAMPLE :. A country’s population in 1994 was 107 million
EXAMPLE : A country’s population in 1994 was 107 million. In 1997 it was 112
million. Estimate the population in 2008 using the exponential
growth model. Round your answer to the nearest million.
P = 107 e14•0.015
P = 107 e0.21
P = 107 • ( )
P =
So the population in 2008 is 132 million.

14. EXPONENTIAL GROWTH
Another area where this model is used is bacterial growth.
where: P = the number of bacteria in the culture at time t
A = the number of bacteria in the culture at time = 0
k = a positive constant of growth
e = the natural logarithm base
t = time in hours

15. EXAMPLE :. A culture started with 5,000 bacteria
EXAMPLE : A culture started with 5,000 bacteria. After 8 hours, it grew to
6,500 bacteria. Predict how many bacteria will be present after
14 hours.
** we will first have to find “k” using the given information
P = 6,500
A = 5,000
t1 = 8 hours
t2 = 14 hours
6,500 = 5,000 e8k

16. EXAMPLE :. A culture started with 5,000 bacteria
EXAMPLE : A culture started with 5,000 bacteria. After 8 hours, it grew to
6,500 bacteria. Predict how many bacteria will be present after
14 hours.
** we will first have to find “k” using the given information
6,500 = 5,000 e8k
1.3 = e8k
Divided both sides by 5,000

17. EXAMPLE :. A culture started with 5,000 bacteria
EXAMPLE : A culture started with 5,000 bacteria. After 8 hours, it grew to
6,500 bacteria. Predict how many bacteria will be present after
14 hours.
** we will first have to find “k” using the given information
6,500 = 5,000 e8k
1.3 = e8k
ln ( 1.3 ) = ln ( e8k )
Take ln of both sides

18. EXAMPLE :. A culture started with 5,000 bacteria
EXAMPLE : A culture started with 5,000 bacteria. After 8 hours, it grew to
6,500 bacteria. Predict how many bacteria will be present after
14 hours.
** we will first have to find “k” using the given information
6,500 = 5,000 e8k
1.3 = e8k
ln ( 1.3 ) = ln ( e8k )
= 8k
Take ln of both sides

19. EXAMPLE :. A culture started with 5,000 bacteria
EXAMPLE : A culture started with 5,000 bacteria. After 8 hours, it grew to
6,500 bacteria. Predict how many bacteria will be present after
14 hours.
** we will first have to find “k” using the given information
6,500 = 5,000 e8k
1.3 = e8k
ln ( 1.3 ) = ln ( e8k )
= 8k
k =
Divide both sides by 8

20. EXAMPLE :. A culture started with 5,000 bacteria
EXAMPLE : A culture started with 5,000 bacteria. After 8 hours, it grew to
6,500 bacteria. Predict how many bacteria will be present after
14 hours.
** we have found k =
Substituted A = 5,000
t2 = 14 hours
k =
P = 5,000 e14 •

21. EXAMPLE :. A culture started with 5,000 bacteria
EXAMPLE : A culture started with 5,000 bacteria. After 8 hours, it grew to
6,500 bacteria. Predict how many bacteria will be present after
14 hours.
P = 5,000 e14 •
P = 5,000 e0.4592
multiplied 14 •

22. EXAMPLE :. A culture started with 5,000 bacteria
EXAMPLE : A culture started with 5,000 bacteria. After 8 hours, it grew to
6,500 bacteria. Predict how many bacteria will be present after
14 hours.
P = 5,000 e14 •
P = 5,000 e0.4592
P = 5,000 • 1.583
evaluated e0.4592

23. EXAMPLE :. A culture started with 5,000 bacteria
EXAMPLE : A culture started with 5,000 bacteria. After 8 hours, it grew to
6,500 bacteria. Predict how many bacteria will be present after
14 hours.
P = 5,000 e14 •
P = 5,000 e0.4592
P = 5,000 • 1.583
P = 7,915 bacteria
multiplied