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68 applications of exponential and log

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  • 1. Applications of Log and Exponential Formulas
  • 2. Applications of Log and Exponential FormulasRecall the following compound interest formulas.Pina (periodic compounding)and Perta (continuous compounding)
  • 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Applications of Log and Exponential FormulasThe constant ln(2) is rounded up to 0.72 for estimation.
  • 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. 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. 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. 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. 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. 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. 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. 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. Applications of Log and Exponential FormulasLikewise if D is known then the growth rate r is In(2) . D
  • 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. 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. 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. 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. 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. 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. 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. Applications of Log and Exponential FormulasIn a similar manner we may define tripling time orquadrupling time,
  • 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. 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. 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. 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. 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. 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. 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.