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- 1. Calculation with log and Exp
- 2. Calculation with log and ExpIn this section, we solve simple numerical equationsinvolving log and exponential functions in base 10or base e.
- 3. Calculation with log and ExpIn this section, we solve simple numerical equationsinvolving log and exponential functions in base 10or base e. Most numerical calculations in science arein these two bases.
- 4. Calculation with log and ExpIn this section, we solve simple numerical equationsinvolving log and exponential functions in base 10or base e. Most numerical calculations in science arein these two bases. We need a calculator that hasthe following functions. ex, 10x, ln(x), and log(x).
- 5. Calculation with log and ExpIn this section, we solve simple numerical equationsinvolving log and exponential functions in base 10or base e. Most numerical calculations in science arein these two bases. We need a calculator that hasthe following functions. ex, 10x, ln(x), and log(x).All answers are given to 3 significant digits.
- 6. Calculation with log and ExpIn this section, we solve simple numerical equationsinvolving log and exponential functions in base 10or base e. Most numerical calculations in science arein these two bases. We need a calculator that hasthe following functions. ex, 10x, ln(x), and log(x).All answers are given to 3 significant digits.Example A. Find the answers with a calculator. 6a.10 3.32 b. √e = e1/6c. log(4.35) d. ln(2/3)
- 7. Calculation with log and ExpIn this section, we solve simple numerical equationsinvolving log and exponential functions in base 10or base e. Most numerical calculations in science arein these two bases. We need a calculator that hasthe following functions. ex, 10x, ln(x), and log(x).All answers are given to 3 significant digits.Example A. Find the answers with a calculator. 6a.103.32 b. √e = e1/6 ≈ 2090c. log(4.35) d. ln(2/3)
- 8. Calculation with log and ExpIn this section, we solve simple numerical equationsinvolving log and exponential functions in base 10or base e. Most numerical calculations in science arein these two bases. We need a calculator that hasthe following functions. ex, 10x, ln(x), and log(x).All answers are given to 3 significant digits.Example A. Find the answers with a calculator. 6a.103.32 b. √e = e1/6 ≈ 2090 ≈ 1.18c. log(4.35) d. ln(2/3)
- 9. Calculation with log and ExpIn this section, we solve simple numerical equationsinvolving log and exponential functions in base 10or base e. Most numerical calculations in science arein these two bases. We need a calculator that hasthe following functions. ex, 10x, ln(x), and log(x).All answers are given to 3 significant digits.Example A. Find the answers with a calculator. 6a.103.32 b. √e = e1/6 ≈ 2090 ≈ 1.18c. log(4.35) d. ln(2/3) ≈0.638
- 10. Calculation with log and ExpIn this section, we solve simple numerical equationsinvolving log and exponential functions in base 10or base e. Most numerical calculations in science arein these two bases. We need a calculator that hasthe following functions. ex, 10x, ln(x), and log(x).All answers are given to 3 significant digits.Example A. Find the answers with a calculator. 6a.103.32 b. √e = e1/6 ≈ 2090 ≈ 1.18c. log(4.35) d. ln(2/3) ≈0.638 ≈ -0.405
- 11. Calculation with log and ExpIn this section, we solve simple numerical equationsinvolving log and exponential functions in base 10or base e. Most numerical calculations in science arein these two bases. We need a calculator that hasthe following functions. ex, 10x, ln(x), and log(x).All answers are given to 3 significant digits.Example A. Find the answers with a calculator. 6a.103.32 b. √e = e1/6 ≈ 2090 ≈ 1.18c. log(4.35) d. ln(2/3) ≈0.638 ≈ -0.405These problems may be stated in alternate forms.
- 12. Calculation with log and ExpExample B. Find the xa. log(x) = 3.32 b. 1/6 = ln(x)c. 10x = 4.35 d. 2/3 = ex
- 13. Calculation with log and ExpExample B. Find the xa. log(x) = 3.32 b. 1/6 = ln(x) x =103.32 (≈ 2090)c. 10x = 4.35 d. 2/3 = ex
- 14. Calculation with log and ExpExample B. Find the xa. log(x) = 3.32 b. 1/6 = ln(x) x =103.32 (≈ 2090) e1/6 = x (≈ 1.18)c. 10x = 4.35 d. 2/3 = ex
- 15. Calculation with log and ExpExample B. Find the xa. log(x) = 3.32 b. 1/6 = ln(x) x =103.32 (≈ 2090) e1/6 = x (≈ 1.18)c. 10x = 4.35 d. 2/3 = ex x = log(4.35) (≈ 0.638)
- 16. Calculation with log and ExpExample B. Find the xa. log(x) = 3.32 b. 1/6 = ln(x) x =103.32 (≈ 2090) e1/6 = x (≈ 1.18)c. 10x = 4.35 d. 2/3 = ex x = log(4.35) (≈ 0.638) ln(2/3) = x (≈ -0.405)
- 17. Calculation with log and ExpExample B. Find the xa. log(x) = 3.32 b. 1/6 = ln(x) x =103.32 (≈ 2090) e1/6 = x (≈ 1.18)c. 10x = 4.35 d. 2/3 = ex x = log(4.35) (≈ 0.638) ln(2/3) = x (≈ -0.405)An equation is called a log-equation if the unknown isin the log-function as in parts a and b above.
- 18. Calculation with log and ExpExample B. Find the xa. log(x) = 3.32 b. 1/6 = ln(x) x =103.32 (≈ 2090) e1/6 = x (≈ 1.18)c. 10x = 4.35 d. 2/3 = ex x = log(4.35) (≈ 0.638) ln(2/3) = x (≈ -0.405)An equation is called a log-equation if the unknown isin the log-function as in parts a and b above.An equation is called an exponential equations if theunknown is in the exponent as in parts c and d.
- 19. Calculation with log and ExpExample B. Find the xa. log(x) = 3.32 b. 1/6 = ln(x) x =103.32 (≈ 2090) e1/6 = x (≈ 1.18)c. 10x = 4.35 d. 2/3 = ex x = log(4.35) (≈ 0.638) ln(2/3) = x (≈ -0.405)An equation is called a log-equation if the unknown isin the log-function as in parts a and b above.An equation is called an exponential equations if theunknown is in the exponent as in parts c and d.To solve log-equations, drop the log and write theproblems in exp-form.
- 20. Calculation with log and ExpExample B. Find the xa. log(x) = 3.32 b. 1/6 = ln(x) x =103.32 (≈ 2090) e1/6 = x (≈ 1.18)c. 10x = 4.35 d. 2/3 = ex x = log(4.35) (≈ 0.638) ln(2/3) = x (≈ -0.405)An equation is called a log-equation if the unknown isin the log-function as in parts a and b above.An equation is called an exponential equations if theunknown is in the exponent as in parts c and d.To solve log-equations, drop the log and write theproblems in exp-form. To solve exponentialequations, lower the exponents and write theproblems in log-form.
- 21. Calculation with log and ExpMore precisely, to solve exponential equations,
- 22. Calculation with log and ExpMore precisely, to solve exponential equations:I. isolate the exponential part that contains the x,
- 23. Calculation with log and ExpMore precisely, to solve exponential equations:I. isolate the exponential part that contains the x,II. bring down the exponents by writing it in log-form.
- 24. Calculation with log and ExpMore precisely, to solve exponential equations:I. isolate the exponential part that contains the x,II. bring down the exponents by writing it in log-form.Example C. Solve 25 = 7*102x
- 25. Calculation with log and ExpMore precisely, to solve exponential equations:I. isolate the exponential part that contains the x,II. bring down the exponents by writing it in log-form.Example C. Solve 25 = 7*102xIsolate the exponential part containing the x,25/7 = 102x
- 26. Calculation with log and ExpMore precisely, to solve exponential equations:I. isolate the exponential part that contains the x,II. bring down the exponents by writing it in log-form.Example C. Solve 25 = 7*102xIsolate the exponential part containing the x,25/7 = 102xBring down the x by restating it in log-form.log(25/7) = 2x
- 27. Calculation with log and ExpMore precisely, to solve exponential equations:I. isolate the exponential part that contains the x,II. bring down the exponents by writing it in log-form.Example C. Solve 25 = 7*102xIsolate the exponential part containing the x,25/7 = 102xBring down the x by restating it in log-form.log(25/7) = 2xlog(25/7) = x 2
- 28. Calculation with log and ExpMore precisely, to solve exponential equations:I. isolate the exponential part that contains the x,II. bring down the exponents by writing it in log-form.Example C. Solve 25 = 7*102xIsolate the exponential part containing the x,25/7 = 102xBring down the x by restating it in log-form.log(25/7) = 2xlog(25/7) = x ≈ 0.276 2
- 29. Calculation with log and ExpMore precisely, to solve exponential equations:I. isolate the exponential part that contains the x,II. bring down the exponents by writing it in log-form.Example C. Solve 25 = 7*102xIsolate the exponential part containing the x,25/7 = 102xBring down the x by restating it in log-form.log(25/7) = 2xlog(25/7) = x ≈ 0.276 2Exact answer Approx. answer
- 30. Calculation with log and ExpExample D. Solve 2.3*e2-3x + 4.1 = 12.5
- 31. Calculation with log and ExpExample D. Solve 2.3*e2-3x + 4.1 = 12.5Isolate the exp-part. 2.3*e2-3x + 4.1 = 12.5
- 32. Calculation with log and ExpExample D. Solve 2.3*e2-3x + 4.1 = 12.5Isolate the exp-part. 2.3*e2-3x + 4.1 = 12.5 2.3*e2-3x = 12.5 – 4.1 2.3*e2-3x = 8.4
- 33. Calculation with log and ExpExample D. Solve 2.3*e2-3x + 4.1 = 12.5Isolate the exp-part. 2.3*e2-3x + 4.1 = 12.5 2.3*e2-3x = 12.5 – 4.1 2.3*e2-3x = 8.4 e2-3x = 8.4/2.3
- 34. Calculation with log and ExpExample D. Solve 2.3*e2-3x + 4.1 = 12.5Isolate the exp-part. 2.3*e2-3x + 4.1 = 12.5 2.3*e2-3x = 12.5 – 4.1 2.3*e2-3x = 8.4 e2-3x = 8.4/2.3Restate in log-form. 2 – 3x = ln(8.4/2.3)
- 35. Calculation with log and ExpExample D. Solve 2.3*e2-3x + 4.1 = 12.5Isolate the exp-part. 2.3*e2-3x + 4.1 = 12.5 2.3*e2-3x = 12.5 – 4.1 2.3*e2-3x = 8.4 e2-3x = 8.4/2.3Restate in log-form. 2 – 3x = ln(8.4/2.3)Solve for x. 2 – ln(8.4/2.3) = 3x
- 36. Calculation with log and ExpExample D. Solve 2.3*e2-3x + 4.1 = 12.5Isolate the exp-part. 2.3*e2-3x + 4.1 = 12.5 2.3*e2-3x = 12.5 – 4.1 2.3*e2-3x = 8.4 e2-3x = 8.4/2.3Restate in log-form. 2 – 3x = ln(8.4/2.3)Solve for x. 2 – ln(8.4/2.3) = 3x 2-ln(8.4/2.3) = x 3
- 37. Calculation with log and ExpExample D. Solve 2.3*e2-3x + 4.1 = 12.5Isolate the exp-part. 2.3*e2-3x + 4.1 = 12.5 2.3*e2-3x = 12.5 – 4.1 2.3*e2-3x = 8.4 e2-3x = 8.4/2.3Restate in log-form. 2 – 3x = ln(8.4/2.3)Solve for x. 2 – ln(8.4/2.3) = 3x 2-ln(8.4/2.3) = x ≈ 0.235 3
- 38. Calculation with log and ExpExample D. Solve 2.3*e2-3x + 4.1 = 12.5Isolate the exp-part. 2.3*e2-3x + 4.1 = 12.5 2.3*e2-3x = 12.5 – 4.1 2.3*e2-3x = 8.4 e2-3x = 8.4/2.3Restate in log-form. 2 – 3x = ln(8.4/2.3)Solve for x. 2 – ln(8.4/2.3) = 3x 2-ln(8.4/2.3) = x ≈ 0.235 3We solve a log-equation in analogous fashion.
- 39. Calculation with log and ExpExample D. Solve 2.3*e2-3x + 4.1 = 12.5Isolate the exp-part. 2.3*e2-3x + 4.1 = 12.5 2.3*e2-3x = 12.5 – 4.1 2.3*e2-3x = 8.4 e2-3x = 8.4/2.3Restate in log-form. 2 – 3x = ln(8.4/2.3)Solve for x. 2 – ln(8.4/2.3) = 3x 2-ln(8.4/2.3) = x ≈ 0.235 3We solve a log-equation in analogous fashion.I. isolate the log part that contains the x,II. drop the log by writing it in exp-form.
- 40. Calculation with log and ExpExample E. Solve 9*log(2x+1)= 7
- 41. Calculation with log and ExpExample E. Solve 9*log(2x+1)= 7Isolate the log-part, log(2x+1) = 7/9
- 42. Calculation with log and ExpExample E. Solve 9*log(2x+1)= 7Isolate the log-part, log(2x+1) = 7/9Write it in exp-form 2x + 1 = 107/9
- 43. Calculation with log and ExpExample E. Solve 9*log(2x+1)= 7Isolate the log-part, log(2x+1) = 7/9Write it in exp-form 2x + 1 = 107/9Save for x.
- 44. Calculation with log and ExpExample E. Solve 9*log(2x+1)= 7Isolate the log-part, log(2x+1) = 7/9Write it in exp-form 2x + 1 = 107/9Save for x. 2x = 107/9 – 1 x = (107/9 – 1)/2
- 45. Calculation with log and ExpExample E. Solve 9*log(2x+1)= 7Isolate the log-part, log(2x+1) = 7/9Write it in exp-form 2x + 1 = 107/9Save for x. 2x = 107/9 – 1 x = (107/9 – 1)/2 ≈ 2.50
- 46. Calculation with log and ExpExample E. Solve 9*log(2x+1)= 7Isolate the log-part, log(2x+1) = 7/9Write it in exp-form 2x + 1 = 107/9Save for x. 2x = 107/9 – 1 x = (107/9 – 1)/2 ≈ 2.50Example F. Solve 2.3*log(2–3x)+4.1 = 12.5
- 47. Calculation with log and ExpExample E. Solve 9*log(2x+1)= 7Isolate the log-part, log(2x+1) = 7/9Write it in exp-form 2x + 1 = 107/9Save for x. 2x = 107/9 – 1 x = (107/9 – 1)/2 ≈ 2.50Example F. Solve 2.3*log(2–3x)+4.1 = 12.52.3*log(2–3x) + 4.1 = 12.5
- 48. Calculation with log and ExpExample E. Solve 9*log(2x+1)= 7Isolate the log-part, log(2x+1) = 7/9Write it in exp-form 2x + 1 = 107/9Save for x. 2x = 107/9 – 1 x = (107/9 – 1)/2 ≈ 2.50Example F. Solve 2.3*log(2–3x)+4.1 = 12.52.3*log(2–3x) + 4.1 = 12.52.3*log(2–3x) = 12.5 – 4.12.3*log(2–3x) = 8.4
- 49. Calculation with log and ExpExample E. Solve 9*log(2x+1)= 7Isolate the log-part, log(2x+1) = 7/9Write it in exp-form 2x + 1 = 107/9Save for x. 2x = 107/9 – 1 x = (107/9 – 1)/2 ≈ 2.50Example F. Solve 2.3*log(2–3x)+4.1 = 12.52.3*log(2–3x) + 4.1 = 12.52.3*log(2–3x) = 12.5 – 4.12.3*log(2–3x) = 8.4log(2 – 3x) = 8.4/2.3
- 50. Calculation with log and ExpExample E. Solve 9*log(2x+1)= 7Isolate the log-part, log(2x+1) = 7/9Write it in exp-form 2x + 1 = 107/9Save for x. 2x = 107/9 – 1 x = (107/9 – 1)/2 ≈ 2.50Example F. Solve 2.3*log(2–3x)+4.1 = 12.52.3*log(2–3x) + 4.1 = 12.52.3*log(2–3x) = 12.5 – 4.12.3*log(2–3x) = 8.4log(2 – 3x) = 8.4/2.3 2 – 3x = 108.4/2.3
- 51. Calculation with log and ExpExample E. Solve 9*log(2x+1)= 7Isolate the log-part, log(2x+1) = 7/9Write it in exp-form 2x + 1 = 107/9Save for x. 2x = 107/9 – 1 x = (107/9 – 1)/2 ≈ 2.50Example F. Solve 2.3*log(2–3x)+4.1 = 12.52.3*log(2–3x) + 4.1 = 12.52.3*log(2–3x) = 12.5 – 4.12.3*log(2–3x) = 8.4log(2 – 3x) = 8.4/2.3 2 – 3x = 108.4/2.3 2 – 108.4/2.3 = 3x
- 52. Calculation with log and ExpExample E. Solve 9*log(2x+1)= 7Isolate the log-part, log(2x+1) = 7/9Write it in exp-form 2x + 1 = 107/9Save for x. 2x = 107/9 – 1 x = (107/9 – 1)/2 ≈ 2.50Example F. Solve 2.3*log(2–3x)+4.1 = 12.52.3*log(2–3x) + 4.1 = 12.52.3*log(2–3x) = 12.5 – 4.12.3*log(2–3x) = 8.4log(2 – 3x) = 8.4/2.3 2 – 3x = 108.4/2.3 2 – 108.4/2.3 = 3x 2 – 108.4/2.3 = x ≈ -1495 3
- 53. Calculation with log and ExpExample G.a. We deposited $5,000 in an account thatcompounds continuously. After 10 years the totalvalue is $18,000. Find the interest rate r.
- 54. Calculation with log and ExpExample G.a. We deposited $5,000 in an account thatcompounds continuously. After 10 years the totalvalue is $18,000. Find the interest rate r.We use the Compound Interest FormulaPert = A with the valuesP = 5,000, A= 18,000 and that t = 10
- 55. Calculation with log and ExpExample G.a. We deposited $5,000 in an account thatcompounds continuously. After 10 years the totalvalue is $18,000. Find the interest rate r.We use the Compound Interest FormulaPert = A with the valuesP = 5,000, A= 18,000 and that t = 10 so5,000e 10r = 18,000
- 56. Calculation with log and ExpExample G.a. We deposited $5,000 in an account thatcompounds continuously. After 10 years the totalvalue is $18,000. Find the interest rate r.We use the Compound Interest FormulaPert = A with the valuesP = 5,000, A= 18,000 and that t = 10 so5,000e 10r = 18,000Hencee10r = 18,000/5,000 = 3.6
- 57. Calculation with log and ExpExample G.a. We deposited $5,000 in an account thatcompounds continuously. After 10 years the totalvalue is $18,000. Find the interest rate r.We use the Compound Interest FormulaPert = A with the valuesP = 5,000, A= 18,000 and that t = 10 so5,000e 10r = 18,000Hencee10r = 18,000/5,000 = 3.6 Apply the natural log to both sides, we have that10r = In (3.6)
- 58. Calculation with log and ExpExample G.a. We deposited $5,000 in an account thatcompounds continuously. After 10 years the totalvalue is $18,000. Find the interest rate r.We use the Compound Interest FormulaPert = A with the valuesP = 5,000, A= 18,000 and that t = 10 so5,000e 10r = 18,000Hencee10r = 18,000/5,000 = 3.6 Apply the natural log to both sides, we have that10r = In (3.6) or r = In (3.6) /10 ≈ 12.8 %.
- 59. Calculation with log and Expb. Write down the specific compound interestformula in the variable t. How much will there be inthe account if we leave the account for 50 years?
- 60. Calculation with log and Expb. Write down the specific compound interestformula in the variable t. How much will there be inthe account if we leave the account for 50 years?We have that P = 5,000, A= 18,000 and r = 0.128.
- 61. Calculation with log and Expb. Write down the specific compound interestformula in the variable t. How much will there be inthe account if we leave the account for 50 years?We have that P = 5,000, A= 18,000 and r = 0.128.Put these value in the Compound Interest FormulaA = Pert
- 62. Calculation with log and Expb. Write down the specific compound interestformula in the variable t. How much will there be inthe account if we leave the account for 50 years?We have that P = 5,000, A= 18,000 and r = 0.128.Put these value in the Compound Interest FormulaA = Pert we get the specific formulaA = 5,000*e0.0128t
- 63. Calculation with log and Expb. Write down the specific compound interestformula in the variable t. How much will there be inthe account if we leave the account for 50 years?We have that P = 5,000, A= 18,000 and r = 0.128.Put these value in the Compound Interest FormulaA = Pert we get the specific formulaA = 5,000*e0.0128tIn 50 years, t = 50,
- 64. Calculation with log and Expb. Write down the specific compound interestformula in the variable t. How much will there be inthe account if we leave the account for 50 years?We have that P = 5,000, A= 18,000 and r = 0.128.Put these value in the Compound Interest FormulaA = Pert we get the specific formulaA = 5,000*e0.0128tIn 50 years, t = 50,we have thatA = 5,000*e0.128 (50)
- 65. Calculation with log and Expb. Write down the specific compound interestformula in the variable t. How much will there be inthe account if we leave the account for 50 years?We have that P = 5,000, A= 18,000 and r = 0.128.Put these value in the Compound Interest FormulaA = Pert we get the specific formulaA = 5,000*e0.0128tIn 50 years, t = 50,we have thatA = 5,000*e0.128 (50)≈ 5,000*e6.4≈ $3,009,225.19
- 66. Calculation with log and ExpGraphically we may view thevalues A in the accountchanges by sliding it alongthe t–axis.
- 67. Calculation with log and ExpGraphically we may view the Avalues A in the accountchanges by sliding it alongthe t–axis. t Graph of A = 5,000*ert, r ≈ 0.128
- 68. Calculation with log and ExpGraphically we may view the Avalues A in the accountchanges by sliding it alongthe t–axis. At t = 0 the initialvalue in the account isA = 5,000 which is alsodenoted as A0. t Graph of A = 5,000*ert, r ≈ 0.128
- 69. Calculation with log and ExpGraphically we may view the Avalues A in the accountchanges by sliding it alongthe t–axis. At t = 0 the initialvalue in the account isA = 5,000 which is also (0, 5,000)denoted as A0. t Graph of A = 5,000*ert, r ≈ 0.128
- 70. Calculation with log and ExpGraphically we may view the Avalues A in the accountchanges by sliding it alongthe t–axis. At t = 0 the initialvalue in the account isA = 5,000 which is also (0, 5,000)denoted as A0. After 50 yrs, tthe account value increases Graph of A = 5,000*ert, r ≈ 0.128to approximate 3,000,000.
- 71. Calculation with log and ExpGraphically we may view the A (50, 3,000,000)values A in the accountchanges by sliding it alongthe t–axis. At t = 0 the initial (not to scale)value in the account isA = 5,000 which is also (0, 5,000)denoted as A0. After 50 yrs, 50 yrs tthe account value increases Graph of A = 5,000*ert, r ≈ 0.128to approximate 3,000,000.
- 72. Calculation with log and ExpGraphically we may view the A (50, 3,000,000)values A in the accountchanges by sliding it alongthe t–axis. At t = 0 the initial (not to scale)value in the account isA = 5,000 which is also (0, 5,000)denoted as A0. After 50 yrs, 50 yrs tthe account value increases Graph of A = 5,000*e , r ≈ 0.128 rtto approximate 3,000,000.Similarly we may set t to be negative to indicate the“past” value in the account.
- 73. Calculation with log and ExpGraphically we may view the A (50, 3,000,000)values A in the accountchanges by sliding it alongthe t–axis. At t = 0 the initial (not to scale)value in the account isA = 5,000 which is also (0, 5,000)denoted as A0. After 50 yrs, 50 yrs tthe account value increases Graph of A = 5,000*e , r ≈ 0.128 rtto approximate 3,000,000.Similarly we may set t to be negative to indicate the“past” value in the account.Hence if t = –5,we have thatA = 5000e0.128 (–5) ≈ 2,636.46.
- 74. Calculation with log and ExpGraphically we may view the A (50, 3,000,000)values A in the accountchanges by sliding it alongthe t–axis. At t = 0 the initial (not to scale)value in the account isA = 5,000 which is also (0, 5,000)denoted as A0. After 50 yrs, (–5, 2636) 50 yrs t 5 yrsthe account value increases Graph of A = 5,000*e , r ≈ 0.128 rtto approximate 3,000,000.Similarly we may set t to be negative to indicate the“past” value in the account.Hence if t = –5,we have thatA = 5000e0.128 (–5) ≈ 2,636.46.
- 75. Calculation with log and ExpExercise A.Solve the following exponential equations, givethe exact and the approximate solutions.1. 5e2x = 7 2. 3e - 2x+1 = 63. 4 – e 3x+ 1 = 2 4. 2* 10 3x - 2 = 55. 6 + 3* 10 1- x = 10 6. -7 – 3*10 2x - 1 = -247. 8 = 12 – 2e 2- x 8. 5*10 2 - 3x + 3 = 14Exercise B. Solve the following log equations, give the exactand the approximate solutions.9. log(3x+1) = 3/5 10. ln(2 – x) = -2/311. 2log(2x –3) = 1/3 12. 2 + log(4 – 2x) = -813. 3 – 5ln(3x +1) = -8 14. -3 +5log(1 – 2x) = 915. 2ln(2x – 1) – 3 = 5 16. 7 – 2ln(12x+15) =23 Exercise C.17. How many years will it take for $1 compoundedcontinuously at the rate r = 4% to double to $2?How about at r = 8%? at r = 12%? at r = 16%?
- 76. Calculation with log and Exp18. In problem 17 would the “doubling times” be different ifthe initial deposit is $1,000, i.e. for the $1,000 to grow to$2,000? Justify your answer.19. In fact the doubling time is t = ln(2)/r which does notdepend on the amount of the initial deposit P. A usefulapproximation for the calculations is to use In(2) ≈ 0.72 so thatt ≈ 0.72 / r. This is called the 72–rule for the doubling times.20. We deposited $1,000 in a account compoundedcontinuously at the rate r. After 20 years the account grew to$20,000. Find r.21. Continue with problem 20, write down the specificcompound interest formula. How much will there be in theaccount if we leave the account for 30 years?22. Continue with problem 20, how much time will it take forthe account to grow to $ I million?
- 77. Calculation with log and Exp23. We deposited $5,000 in a account compoundedcontinuously at the rate r. After 15 years the account grew to$12,000. Find r.24. Continue with problem 23, write down the specificcompound interest formula. How much will there be in theaccount if we leave the account for 100 years?25. Continue with problem 23, how much time will it take forthe account to grow to $ I million?26. We deposited $P in an account compounded continuouslyat r = 5½%. After 5 years there is $5,000 in the account. FindP. Continue with problem 26, write down the specific27.compound interest formula. How much will there be in theaccount if we leave the account for 100 years?28. Continue with problem 26, how much time will it take forthe account to grow to $ I million?
- 78. Calculation with log and ExpSolve the following exponential equations, give the exactand the approximate solutions.1. 5e2x = 7 2. 3e - 2x+1 = 6Exact. x = ½* ln(7/5) Exact. x = (1 – ln(2)) /2Aproxímate. 0.168 Aproxímate. 0.1533. 4 – e 3x+ 1 = 2 4. 2* 10 3x - 2 = 5Exact. x = (ln(2) – 1)/3 Exact. x = (log(5/2) + 2)/3Approximate. - 0.102 Approximate. 0.7995. 6 + 3* 10 1- x = 10 6. -7 – 3*10 2x - 1 = -24Exact. x = 1 – log(4/3) Exact. x = (log(17/3)+1)/2Aproxímate. 0.875 Aproxímate. 0.877
- 79. Solve the following log equations, give the exact and theapproximate solutions.9. log(3x+1) = 3/5 10. ln(2 – x) = -2/3Exact. x = (103/5 – 1)/3 Exact. x = 2 – e -2/3Approximate. 0.994 Approximate. 1.4911. 2log(2x –3) = 1/3 12. 2 + log(4 – 2x) = -8Exact. x = (101/6 + 3)/2 Exact. x = (4 – 10-10)/2Approximate. 2.23 Approximate. 2.00013. 3 – 5ln(3x +1) = -8 14. -3 +5log(1 – 2x) = 9Exact. x = (e11/5 – 1 )/3 Exact. x = (1 – 10 12/5)/2Approximate. 2.68 Approximate. -12515. 2ln(2x – 1) – 3 = 5 16. 7 – 2ln(12x+15) =23

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