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68 applications of exponential and log

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68 applications of exponential and log

1. 1. Applications of Log and Exponential Formulas
2. 2. Applications of Log and Exponential FormulasRecall the following compound interest formulas.Pina (periodic compounding)and Perta (continuous compounding)
3. 3. Applications of Log and Exponential FormulasRecall the following compound interest formulas.Pina (periodic compounding)P = principal, i = periodic rate,N = total number of periods, A = accumulated valueand Perta (continuous compounding)
4. 4. Applications of Log and Exponential FormulasRecall the following compound interest formulas.Pina (periodic compounding)P = principal, i = periodic rate,N = total number of periods, A = accumulated valuethen P(1 + i )N = Aand Perta (continuous compounding)
5. 5. Applications of Log and Exponential FormulasRecall the following compound interest formulas.Pina (periodic compounding)P = principal, i = periodic rate,N = total number of periods, A = accumulated valuethen P(1 + i )N = Aand Perta (continuous compounding)P = principal, r = yearly or annual rate,t = total number of years, A = accumulated value
6. 6. Applications of Log and Exponential FormulasRecall the following compound interest formulas.Pina (periodic compounding)P = principal, i = periodic rate,N = total number of periods, A = accumulated valuethen P(1 + i )N = Aand Perta (continuous compounding)P = principal, r = yearly or annual rate,t = total number of years, A = accumulated valuethen Pert = A
7. 7. Applications of Log and Exponential FormulasRecall the following compound interest formulas.Pina (periodic compounding)P = principal, i = periodic rate,N = total number of periods, A = accumulated valuethen P(1 + i )N = Aand Perta (continuous compounding)P = principal, r = yearly or annual rate,t = total number of years, A = accumulated valuethen Pert = AThe Pina formula has the base (1+ i) which varieswith i.
8. 8. Applications of Log and Exponential FormulasRecall the following compound interest formulas.Pina (periodic compounding)P = principal, i = periodic rate,N = total number of periods, A = accumulated valuethen P(1 + i )N = Aand Perta (continuous compounding)P = principal, r = yearly or annual rate,t = total number of years, A = accumulated valuethen Pert = AThe Pina formula has the base (1+ i) which varieswith i. All the discussions below are related tocontinuous–compounding growth which utilizes thePerta formula with the fixed natural base e.
9. 9. Applications of Log and Exponential FormulasExample A. a. An investment gives 4% annual returncompounded continuously. How long would it takefor an initial investment of \$500 to reach \$1,000?
10. 10. Applications of Log and Exponential FormulasExample A. a. An investment gives 4% annual returncompounded continuously. How long would it takefor an initial investment of \$500 to reach \$1,000?We are to find t using the Perta formula withP = \$500, A= \$1,000, and r = 4% = 0.04.
11. 11. Applications of Log and Exponential FormulasExample A. a. An investment gives 4% annual returncompounded continuously. How long would it takefor an initial investment of \$500 to reach \$1,000?We are to find t using the Perta formula withP = \$500, A= \$1,000, and r = 4% = 0.04. So 500e0.04t = 1000
12. 12. Applications of Log and Exponential FormulasTo find the r or t in the formula Pert = A, divide bothsides by P.Example A. a. An investment gives 4% annual returncompounded continuously. How long would it takefor an initial investment of \$500 to reach \$1,000?We are to find t using the Perta formula withP = \$500, A= \$1,000, and r = 4% = 0.04. So 500e0.04t = 1000
13. 13. Applications of Log and Exponential FormulasTo find the r or t in the formula Pert = A, divide bothsides by P.Example A. a. An investment gives 4% annual returncompounded continuously. How long would it takefor an initial investment of \$500 to reach \$1,000?We are to find t using the Perta formula withP = \$500, A= \$1,000, and r = 4% = 0.04. So 500e0.04t = 1000 e0.04t = 2 (=1000/500)
14. 14. Applications of Log and Exponential FormulasTo find the r or t in the formula Pert = A, divide bothsides by P. We get ert = A so that r*t = ln( A ). P PExample A. a. An investment gives 4% annual returncompounded continuously. How long would it takefor an initial investment of \$500 to reach \$1,000?We are to find t using the Perta formula withP = \$500, A= \$1,000, and r = 4% = 0.04. So 500e0.04t = 1000 e0.04t = 2 (=1000/500)
15. 15. Applications of Log and Exponential FormulasTo find the r or t in the formula Pert = A, divide bothsides by P. We get ert = A so that r*t = ln( A ). P PExample A. a. An investment gives 4% annual returncompounded continuously. How long would it takefor an initial investment of \$500 to reach \$1,000?We are to find t using the Perta formula withP = \$500, A= \$1,000, and r = 4% = 0.04. So 500e0.04t = 1000 e0.04t = 2 (=1000/500) in the log–form 0.04t = ln(2)
16. 16. Applications of Log and Exponential FormulasTo find the r or t in the formula Pert = A, divide bothsides by P. We get ert = A so that r*t = ln( A ). P PExample A. a. An investment gives 4% annual returncompounded continuously. How long would it takefor an initial investment of \$500 to reach \$1,000?We are to find t using the Perta formula withP = \$500, A= \$1,000, and r = 4% = 0.04. So 500e0.04t = 1000 e0.04t = 2 (=1000/500) in the log–form 0.04t = ln(2) or t = ln(2)/0.04 ≈ 17.5 year
17. 17. Applications of Log and Exponential FormulasTo find the r or t in the formula Pert = A, divide bothsides by P. We get ert = A so that r*t = ln( A ). P PExample A. a. An investment gives 4% annual returncompounded continuously. How long would it takefor an initial investment of \$500 to reach \$1,000?We are to find t using the Perta formula withP = \$500, A= \$1,000, and r = 4% = 0.04. So 500e0.04t = 1000 e0.04t = 2 (=1000/500) in the log–form 0.04t = ln(2) or t = ln(2)/0.04 ≈ 17.5 yearSo it would take roughly than 17½ years for the \$500investment to reach \$1,000.
18. 18. Applications of Log and Exponential Formulasb. As before with r = 4%, does it take more time foran investment of \$500 to become \$1,000than for an investment of \$1 to become \$2?
19. 19. Applications of Log and Exponential Formulasb. As before with r = 4%, does it take more time foran investment of \$500 to become \$1,000than for an investment of \$1 to become \$2?No. By viewing the \$500 investment as 500 separate\$1 accounts, we see that the time needed for\$500 to grow into \$1,000 is the same the time neededfor \$1 to grow into \$2
20. 20. Applications of Log and Exponential Formulasb. As before with r = 4%, does it take more time foran investment of \$500 to become \$1,000than for an investment of \$1 to become \$2?No. By viewing the \$500 investment as 500 separate\$1 accounts, we see that the time needed for\$500 to grow into \$1,000 is the same the time neededfor \$1 to grow into \$2 which is ln(2)/0.04 ≈ 17.5 years.
21. 21. Applications of Log and Exponential Formulasb. As before with r = 4%, does it take more time foran investment of \$500 to become \$1,000than for an investment of \$1 to become \$2?No. By viewing the \$500 investment as 500 separate\$1 accounts, we see that the time needed for\$500 to grow into \$1,000 is the same the time neededfor \$1 to grow into \$2 which is ln(2)/0.04 ≈ 17.5 years.Doubling Time and the Rule of 72
22. 22. Applications of Log and Exponential Formulasb. As before with r = 4%, does it take more time foran investment of \$500 to become \$1,000than for an investment of \$1 to become \$2?No. By viewing the \$500 investment as 500 separate\$1 accounts, we see that the time needed for\$500 to grow into \$1,000 is the same the time neededfor \$1 to grow into \$2 which is ln(2)/0.04 ≈ 17.5 years.Doubling Time and the Rule of 72From the above reasoning we see that given a fixedgrowth rate r, the amount of time needed to double insize is the same regardless of the initial size.
23. 23. Applications of Log and Exponential Formulasb. As before with r = 4%, does it take more time foran investment of \$500 to become \$1,000than for an investment of \$1 to become \$2?No. By viewing the \$500 investment as 500 separate\$1 accounts, we see that the time needed for\$500 to grow into \$1,000 is the same the time neededfor \$1 to grow into \$2 which is ln(2)/0.04 ≈ 17.5 years.Doubling Time and the Rule of 72From the above reasoning we see that given a fixedgrowth rate r, the amount of time needed to double insize is the same regardless of the initial size.The amount of time D that’s needed for doubling insize is called the “doubling time (at rate r)”.
24. 24. Applications of Log and Exponential Formulasb. As before with r = 4%, does it take more time foran investment of \$500 to become \$1,000than for an investment of \$1 to become \$2?No. By viewing the \$500 investment as 500 separate\$1 accounts, we see that the time needed for\$500 to grow into \$1,000 is the same the time neededfor \$1 to grow into \$2 which is ln(2)/0.04 ≈ 17.5 years.Doubling Time and the Rule of 72From the above reasoning we see that given a fixedgrowth rate r, the amount of time needed to double insize is the same regardless of the initial size.The amount of time D that’s needed for doubling insize is called the “doubling time (at rate r)”.So at r = 4%, the doubling time D ≈ 17.5 yrs.
25. 25. Applications of Log and Exponential FormulasIn example A. the doubling time D at r = 0.04 isD = In(2) ≈ 17.5 yrs. 0.04
26. 26. Applications of Log and Exponential FormulasIn example A. the doubling time D at r = 0.04 isD = In(2) ≈ 17.5 yrs. 0.04Similarly if r = 5% = 0.05, we solve for t in the equatione0.05t = ln(2)
27. 27. Applications of Log and Exponential FormulasIn example A. the doubling time D at r = 0.04 isD = In(2) ≈ 17.5 yrs. 0.04Similarly if r = 5% = 0.05, we solve for t in the equation In(2)e0.05t = ln(2) so the doubling time is 0.05 ≈ 14.0 yrs.
28. 28. Applications of Log and Exponential FormulasIn example A. the doubling time D at r = 0.04 isD = In(2) ≈ 17.5 yrs. 0.04Similarly if r = 5% = 0.05, we solve for t in the equation In(2)e0.05t = ln(2) so the doubling time is 0.05 ≈ 14.0 yrs.Leaving the rate r as a variable, the same algebragives us the following.The Doubling Time Formula
29. 29. Applications of Log and Exponential FormulasIn example A. the doubling time D at r = 0.04 isD = In(2) ≈ 17.5 yrs. 0.04Similarly if r = 5% = 0.05, we solve for t in the equation In(2)e0.05t = ln(2) so the doubling time is 0.05 ≈ 14.0 yrs.Leaving the rate r as a variable, the same algebragives us the following.The Doubling Time FormulaFor the exponential growth with rate r, i.e. Pe rt = A,the doubling time D = r In(2) ( ≈ 0.69897 ) r .Proof.
30. 30. Applications of Log and Exponential FormulasIn example A. the doubling time D at r = 0.04 isD = In(2) ≈ 17.5 yrs. 0.04Similarly if r = 5% = 0.05, we solve for t in the equation In(2)e0.05t = ln(2) so the doubling time is 0.05 ≈ 14.0 yrs.Leaving the rate r as a variable, the same algebragives us the following.The Doubling Time FormulaFor the exponential growth with rate r, i.e. Pe rt = A,the doubling time D = r In(2) ( ≈ 0.69897 ) r .Proof. To double means to have an initial amount ofP = 1 unit to grow into A = 2 units.
31. 31. Applications of Log and Exponential FormulasIn example A. the doubling time D at r = 0.04 isD = In(2) ≈ 17.5 yrs. 0.04Similarly if r = 5% = 0.05, we solve for t in the equation In(2)e0.05t = ln(2) so the doubling time is 0.05 ≈ 14.0 yrs.Leaving the rate r as a variable, the same algebragives us the following.The Doubling Time FormulaFor the exponential growth with rate r, i.e. Pe rt = A,the doubling time D = r In(2) ( ≈ 0.69897 ) r .Proof. To double means to have an initial amount ofP = 1 unit to grow into A = 2 units. Hence 1* er*t = 2.
32. 32. Applications of Log and Exponential FormulasIn example A. the doubling time D at r = 0.04 isD = In(2) ≈ 17.5 yrs. 0.04Similarly if r = 5% = 0.05, we solve for t in the equation In(2)e0.05t = ln(2) so the doubling time is 0.05 ≈ 14.0 yrs.Leaving the rate r as a variable, the same algebragives us the following.The Doubling Time FormulaFor the exponential growth with rate r, i.e. Pe rt = A,the doubling time D = r In(2) ( ≈ 0.69897 ) r .Proof. To double means to have an initial amount ofP = 1 unit to grow into A = 2 units. Hence 1* er*t = 2.So r * t = ln(2)
33. 33. Applications of Log and Exponential FormulasIn example A. the doubling time D at r = 0.04 isD = In(2) ≈ 17.5 yrs. 0.04Similarly if r = 5% = 0.05, we solve for t in the equation In(2)e0.05t = ln(2) so the doubling time is 0.05 ≈ 14.0 yrs.Leaving the rate r as a variable, the same algebragives us the following.The Doubling Time FormulaFor the exponential growth with rate r, i.e. Pe rt = A,the doubling time D = r In(2) ( ≈ 0.69897 ) r .Proof. To double means to have an initial amount ofP = 1 unit to grow into A = 2 units. Hence 1* er*t = 2.So r * t = ln(2) or t = ln(2)/r = D is the doubling time.
34. 34. Applications of Log and Exponential FormulasThe constant ln(2) is rounded up to 0.72 for estimation.
35. 35. Applications of Log and Exponential FormulasThe constant ln(2) is rounded up to 0.72 for estimation.The Rule of 72 for the Doubling TimeThe doubling time D ≈ 0.72 where r is the growth rate. r
36. 36. Applications of Log and Exponential FormulasThe constant ln(2) is rounded up to 0.72 for estimation.The Rule of 72 for the Doubling TimeThe doubling time D ≈ 0.72 where r is the growth rate. rWe use 0.72 because 72 can be divided by manydifferent values, hence convenient to use.
37. 37. Applications of Log and Exponential FormulasThe constant ln(2) is rounded up to 0.72 for estimation.The Rule of 72 for the Doubling TimeThe doubling time D ≈ 0.72 where r is the growth rate. rWe use 0.72 because 72 can be divided by manydifferent values, hence convenient to use.So at r = 4%, the doubling time ≈ 0.72 = 72 = 18 0.04 4 yrs,
38. 38. Applications of Log and Exponential FormulasThe constant ln(2) is rounded up to 0.72 for estimation.The Rule of 72 for the Doubling TimeThe doubling time D ≈ 0.72 where r is the growth rate. rWe use 0.72 because 72 can be divided by manydifferent values, hence convenient to use.So at r = 4%, the doubling time ≈ 0.72 = 72 = 18 0.04 4if r = 8%, the doubling time ≈ 72 = 9 yrs, 8 yrs,
39. 39. Applications of Log and Exponential FormulasThe constant ln(2) is rounded up to 0.72 for estimation.The Rule of 72 for the Doubling TimeThe doubling time D ≈ 0.72 where r is the growth rate. rWe use 0.72 because 72 can be divided by manydifferent values, hence convenient to use.So at r = 4%, the doubling time ≈ 0.72 = 72 = 18 0.04 4if r = 8%, the doubling time ≈ 72 = 9 yrs, 8 yrs, 72if r = 12%, the doubling time ≈ 12 = 6 yrs.
40. 40. Applications of Log and Exponential FormulasThe constant ln(2) is rounded up to 0.72 for estimation.The Rule of 72 for the Doubling TimeThe doubling time D ≈ 0.72 where r is the growth rate. rWe use 0.72 because 72 can be divided by manydifferent values, hence convenient to use.So at r = 4%, the doubling time ≈ 0.72 = 72 = 18 0.04 4if r = 8%, the doubling time ≈ 72 = 9 yrs, 8 yrs, 72if r = 12%, the doubling time ≈ 12 = 6 yrs.The equation D = In(2) sayrs that “the doubling time rD is inversely proportional to the growth rate r”.
41. 41. Applications of Log and Exponential FormulasThe constant ln(2) is rounded up to 0.72 for estimation.The Rule of 72 for the Doubling TimeThe doubling time D ≈ 0.72 where r is the growth rate. rWe use 0.72 because 72 can be divided by manydifferent values, hence convenient to use.So at r = 4%, the doubling time ≈ 0.72 = 72 = 18 0.04 4if r = 8%, the doubling time ≈ 72 = 9 yrs, 8 yrs, 72if r = 12%, the doubling time ≈ 12 = 6 yrs.The equation D = In(2) sayrs that “the doubling time rD is inversely proportional to the growth rate r”.So if we double the growth rate r then D is shorten to1/2 of the time before,
42. 42. Applications of Log and Exponential Formulas The constant ln(2) is rounded up to 0.72 for estimation. The Rule of 72 for the Doubling Time The doubling time D ≈ 0.72 where r is the growth rate. r We use 0.72 because 72 can be divided by many different values, hence convenient to use. So at r = 4%, the doubling time ≈ 0.72 = 72 = 18 0.04 4 if r = 8%, the doubling time ≈ 72 = 9 yrs, 8 yrs, 72 if r = 12%, the doubling time ≈ 12 = 6 yrs.The equation D = In(2) sayrs that “the doubling time rD is inversely proportional to the growth rate r”.So if we double the growth rate r then D is shorten to1/2 of the time before, and if we triple (3x) the rate rthen D is shorten to 1/3 as before as shown above.
43. 43. Applications of Log and Exponential FormulasLikewise if D is known then the growth rate r is In(2) . D
44. 44. Applications of Log and Exponential FormulasLikewise if D is known then the growth rate r is In(2) . DThe Rule of 72 for the Growth RateGiven the doubling time D, the approximate interestrate r ≈ 0.72 D
45. 45. Applications of Log and Exponential FormulasLikewise if D is known then the growth rate r is In(2) . DThe Rule of 72 for the Growth RateGiven the doubling time D, the approximate interestrate r ≈ 0.72 DExample B. What is the approximate interest rate r ifthe doubling time D = 24 yrs? D =12 yrs? D= 6 yrs?
46. 46. Applications of Log and Exponential FormulasLikewise if D is known then the growth rate r is In(2) . DThe Rule of 72 for the Growth RateGiven the doubling time D, the approximate interestrate r ≈ 0.72 DExample B. What is the approximate interest rate r ifthe doubling time D = 24 yrs? D =12 yrs? D= 6 yrs?Using the Rule of 72, if D = 24 yrs, then r ≈0.72 = 3% 24
47. 47. Applications of Log and Exponential FormulasLikewise if D is known then the growth rate r is In(2) . DThe Rule of 72 for the Growth RateGiven the doubling time D, the approximate interestrate r ≈ 0.72 DExample B. What is the approximate interest rate r ifthe doubling time D = 24 yrs? D =12 yrs? D= 6 yrs?Using the Rule of 72, if D = 24 yrs, then r ≈0.72 = 3% 24If D = 12 yrs, then r ≈ 0.72 = 6% 12
48. 48. Applications of Log and Exponential FormulasLikewise if D is known then the growth rate r is In(2) . DThe Rule of 72 for the Growth RateGiven the doubling time D, the approximate interestrate r ≈ 0.72 DExample B. What is the approximate interest rate r ifthe doubling time D = 24 yrs? D =12 yrs? D= 6 yrs?Using the Rule of 72, if D = 24 yrs, then r ≈0.72 = 3% 24If D = 12 yrs, then r ≈ 0.72 = 6% 12If D = 8 yrs, then r ≈0.72 = 9% 8
49. 49. Applications of Log and Exponential FormulasLikewise if D is known then the growth rate r is In(2) . DThe Rule of 72 for the Growth RateGiven the doubling time D, the approximate interestrate r ≈ 0.72 DExample B. What is the approximate interest rate r ifthe doubling time D = 24 yrs? D =12 yrs? D= 6 yrs?Using the Rule of 72, if D = 24 yrs, then r ≈0.72 = 3% 24If D = 12 yrs, then r ≈ 0.72 = 6% 12If D = 8 yrs, then r ≈0.72 = 9% 8So if D is shorten to 1/2 of the time before, then thegrowth rate r must be doubled,
50. 50. Applications of Log and Exponential FormulasLikewise if D is known then the growth rate r is In(2) . DThe Rule of 72 for the Growth RateGiven the doubling time D, the approximate interestrate r ≈ 0.72 DExample B. What is the approximate interest rate r ifthe doubling time D = 24 yrs? D =12 yrs? D= 6 yrs?Using the Rule of 72, if D = 24 yrs, then r ≈0.72 = 3% 24If D = 12 yrs, then r ≈ 0.72 = 6% 12If D = 8 yrs, then r ≈0.72 = 9% 8So if D is shorten to 1/2 of the time before, then thegrowth rate r must be doubled, and if D is to be shortento 1/3 as before, then the growth rate r must be tripled.
51. 51. Applications of Log and Exponential FormulasIn a similar manner we may define tripling time orquadrupling time,
52. 52. Applications of Log and Exponential FormulasIn a similar manner we may define tripling time orquadrupling time, i.e. the time it takes for the growth tobecome three times or four times of its original size.
53. 53. Applications of Log and Exponential FormulasIn a similar manner we may define tripling time orquadrupling time, i.e. the time it takes for the growth tobecome three times or four times of its original size.The tripling time = In(3) ≈ 1.08 r r
54. 54. Applications of Log and Exponential FormulasIn a similar manner we may define tripling time orquadrupling time, i.e. the time it takes for the growth tobecome three times or four times of its original size.The tripling time = In(3) ≈ 1.08 r rThe quadrupling time = In(4) ≈ 1.44 r r
55. 55. Applications of Log and Exponential FormulasIn a similar manner we may define tripling time orquadrupling time, i.e. the time it takes for the growth tobecome three times or four times of its original size.The tripling time = In(3) ≈ 1.08 r rThe quadrupling time = In(4) ≈ 1.44 r rBy the same algebra for deriving the doubling timewe obtain the following from the Perta formula.
56. 56. Applications of Log and Exponential FormulasIn a similar manner we may define tripling time orquadrupling time, i.e. the time it takes for the growth tobecome three times or four times of its original size.The tripling time = In(3) ≈ 1.08 r rThe quadrupling time = In(4) ≈ 1.44 r rBy the same algebra for deriving the doubling timewe obtain the following from the Perta formula.It takes In(K) years to grow to K times of the original rsize with continuous growth rate r.
57. 57. Applications of Log and Exponential FormulasIn a similar manner we may define tripling time orquadrupling time, i.e. the time it takes for the growth tobecome three times or four times of its original size.The tripling time = In(3) ≈ 1.08 r rThe quadrupling time = In(4) ≈ 1.44 r rBy the same algebra for deriving the doubling timewe obtain the following from the Perta formula.It takes In(K) years to grow to K times of the original rsize with continuous growth rate r. In(10So it would takes 0.08 ≈ 27.8 years at the continuous )rate 8% to grow to to 10 times of the original amount.
58. 58. Applications of Log and Exponential FormulasThe variable K is called the factor.• For the exponential growth (r > 0), the factor K isindependent of the initial size.• It would take In(K)/r yrs to reach the size of factor K.If r < 0 then we have a process of exponential decay,contraction, depreciation, etc.. and as time goes on,the factor K is less than 1. For example, if r = –0.05then after 10 year, the initial unit would shrink to thefactor K = e–0.05(10) = e–0.5 ≈ 3/5 of the original.• For the exponential decay (r < 0), the factor K isindependent of the initial size.• It would take In(K)/r yrs to reach the size of factor K.