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

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.

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 )

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

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 )

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

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”…

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

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

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 • ( 1.234 ) P = 132.004 Multiplied 14 times 0.015

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 • ( 1.234 ) P = 132.004 - Evaluated e0.165

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 • ( 1.234 ) P = 132.004 - multiplied

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 • ( 1.234 ) P = 132.004 So the population in 2008 is 132 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

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

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

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

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 ) 0.2624 = 8k Take ln of both sides

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 ) 0.2624 = 8k k = 0.0328 Divide both sides by 8

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 = 0.0328 Substituted A = 5,000 t2 = 14 hours k = 0.0328 P = 5,000 e14 • 0.0328

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 • 0.0328 P = 5,000 e0.4592 multiplied 14 • 0.0328

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 • 0.0328 P = 5,000 e0.4592 P = 5,000 • 1.583 evaluated e0.4592

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 • 0.0328 P = 5,000 e0.4592 P = 5,000 • 1.583 P = 7,915 bacteria multiplied