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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
ktAeP
ktAeP
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.
ktAeP
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 )
ktAeP
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
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 )
ke3107112
ktAeP
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.
k
k
k
e
e
e
3
3
3
047.1
107
107
107
112
107112
- divide both sides by 107
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.
0153.3
3
3
046.
3046.
ln)047.1ln(
047.1
107
107
107
112
107112
3
3
3
3
k
k
k
e
e
e
e
k
k
k
k
- divide both sides by 107
- take “ln” by both sides
This gives us “k”…
ktAeP
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
ktAeP
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
ktAeP
Multiplied 14 times 0.015
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 • ( 1.234 )
P = 132.004
ktAeP
- Evaluated e0.165
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 • ( 1.234 )
P = 132.004
ktAeP
- multiplied
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 • ( 1.234 )
P = 132.004
ktAeP
P = 107 e14•0.015
P = 107 e0.21
P = 107 • ( 1.234 )
P = 132.004
So the population in 2008 is 132 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.
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
ktAeP
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.
ktAeP P = 6,500
A = 5,000
t1 = 8 hours
t2 = 14 hours
** we will first have to find “k” using the given information
6,500 = 5,000 e8k
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.
ktAeP
Divided both sides by 5,000
** we will first have to find “k” using the given information
6,500 = 5,000 e8k
1.3 = e8k
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.
ktAeP
** 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
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.
ktAeP
** we will first have to find “k” using the given information
6,500 = 5,000 e8k
1.3 = e8k
ln ( 1.3 ) = ln ( e8k )
0.2624 = 8k
Take ln of both sides
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.
ktAeP
** we will first have to find “k” using the given information
6,500 = 5,000 e8k
1.3 = e8k
ln ( 1.3 ) = ln ( e8k )
0.2624 = 8k
k = 0.0328
Divide both sides by 8
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.
ktAeP
** we have found k = 0.0328
P = 5,000 e14 • 0.0328 Substituted A = 5,000
t2 = 14 hours
k = 0.0328
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.
ktAeP P = 5,000 e14 • 0.0328
P = 5,000 e0.4592 multiplied 14 • 0.0328
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.
ktAeP P = 5,000 e14 • 0.0328
P = 5,000 e0.4592
P = 5,000 • 1.583 evaluated e0.4592
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.
ktAeP P = 5,000 e14 • 0.0328
P = 5,000 e0.4592
P = 5,000 • 1.583
P = 7,915 bacteria multiplied
