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EXPONENTIAL GROWTH

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

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EXPONENTIAL GROWTH

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  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. 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. 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. 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. 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. 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. 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. 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. 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. 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

  11. 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

  12. 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

  13. 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.

  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. 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. 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. 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. 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

  19. 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

  20. 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

  21. 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

  22. 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

  23. 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

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