exponential growth exponential functions can be applied to real – world problems. one instance...

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