4.6 Eponential and Logarithmic Equations (Part I) In this section you will learn to: solve eponential equations using like ases solve eponential equations using logarithms solve logarithmic equations using the definition of a logarithm solve logarithmic equations using 1-to-1 properties of logarithms apply logarithmic and eponential equations to real-world prolems convert y = a to an eponential equation using ase e Definition of a Logarithm y = log is equivalent to y = Inverse Properties log log = = Log Properties Involving One log = 1 log 1 = 0 Product Rule log ( MN) = log M + log N Quotient Rule log M N = log M log N Power Rule p log M = p log M If M N = then M = N. One-to-One Properties If log M = log N then M = N. If M = N then log M = log N. Page 1 (Section 4.6)
Eample 1: Solve each equation y epressing each side as a power of the same ase. 1 3 (a) 5 + 1 = 5 () 9 = (c) e e 5 3 = e e 6 Steps for solving EXPONENTIAL EQUATIONS: Eample : Solve 5e = 60 (Eamples 6) 1.. 3. 4. Eample 3: Solve 3 = 30 using (a) common logarithms, () natural logarithms, and (c) the definition of a logarithm. Page (Section 4.6)
Eample 4: Solve 10 + 3 = 835 Eample 5: Solve 5 = 50 Eample 6: Solve + 1 = 3 =========================================================================== Eample 7: Use FACTORING to solve each of the following equations. (Hint: Use sustitution or short-cut method learned in Section 1.6.) (a) e e 3 = 0 () 3 4 3 1 = 0 Page 3 (Section 4.6)
=========================================================================== Steps for solving LOGARITHMIC EQUATIONS: Eample 8: Solve log 4 ( + 3) = (Eamples 8 11) 1.. 3. Eample 9: Solve log + log ( + 7) = 3 Eample 10: Solve 3 ln = 1 Eample 11: Solve log ( + ) log ( 5) = 1 =========================================================================== Steps for solving equations using 1-to-1 properties: Eample 1: log( + 7) log3 = log(7 + 1) (Eamples 1 14) 1.. 3. Page 4 (Section 4.6)
Eample 13: log log7 = log11 Eample 14: ln( 3) = ln(7 3) ln( + 1) =========================================================================== Periodic Interest Formula Continuous Interest Formula nt r A = P 1 + rt n A = Pe Eample 15: How long will it take $5,000 Eample 16: How long will it take $5,000 to grow to $500,000 at 9% interest to grow to $500,000 at 9% interest compounded continuously? compounded quarterly? Eample 17: What interest rate is needed for $5,000 to doule after 8 years if compounded continuously? (Round rate to nearest hundredth of a percent.) Page 5 (Section 4.6)
Part II: Eponential Growth and Decay Decay Model rt A = Pe ( t) = A e f 0 A = A0e kt kt Growth Model A 0 = A = k = t = Eample 18: In 001 the world population was approimately 6. illion. If the annual growth rate averaged aout 1.3% per year, write an eponential equation that models this situation. Use your model to estimate the population for this year. Eample 19: An account has a continuous interest rate of k. (a) How long will it take your money to doule if compounded continuously? () How long will it take it to triple? (c) At 3% interest, how long will it take an investment to doule? Triple? (d) What interest rate is needed for an investment to doule after 5 years? Page 6 (Section 4.6)
Steps for finding growth/decay model (when growth or decay rate is not given: 1. Use kt A = A0e to find k. (Estimate k to 6 decimal places.). Sustitute k into growth/decay model: A 0 = A e kt Eample 0: The minimum wage in 1970 was $1.60. In 000 it was $5.15. (a) Find a growth model for this situation. () Estimate the minimum wage for this year. (c) Estimate the minimum wage 10 years from now. (d) Based on this model, when will the minimum wage e $10.00? Page 7 (Section 4.6)
Eample 1: Caron-14 testing is used to determine the age of fossils, artifacts and paintings. Caron- 14 has a half-life of 5715 years. (a) Find an eponential decay model for Caron-14. () A painting was discovered containing 96% of its original Caron-14. Estimate the age of the painting. (c) An art collector plans to purchase a painting y Leonardo DaVinci for a considerale amount of money. (DaVinci lived from 145-1519). Could the painting in part () possily e one of DaVinci s works? Page 8 (Section 4.6)
Eample : Arsenic-74 has a half-life of 17.5 days. (a) Find an eponential decay model for this situation. () How long will it take for 5% of the original amount of arsenic-74 to remain in a lood system? (Round to nearest day.) (c) What is the decay rate (per day) to the nearest tenth of a percent. Epressing an Eponential Model in Base e: y = ae ln y = a is equivalent to y (ln ) = ae or Eample 3: Rewrite each equation in terms of ase e. (a) y 68() = () y 5000(.5) = (c) y.7(0.5) = (d) y = 5(3.5) Page 9 (Section 4.6)
4.6 Homework Prolems Solve each equation y epressing each side as a power of the same ase. 1. 3 + 3 = 9. 5 = 3 1 5 7 e 3. e e = 4. e Solve the eponential equations. Round answers to two decimal places. 1 3 + 1 3 = 9 5. 5 = 7 6. 3 e = 30 7. 10 + 36 = 150 8. 5 3 = 137 4 5 9. e 7 = 11, 43 10. 5 + 3 1 = 3 11. e + 1 = 10 1. 40 0. 6 e 3 = 37 Solve y factoring. Write the answer using eact values. 13. e 4e + 3 = 0 14. e 3e + = 0 15. e e = 3 16. 4 5 e + e = 4 Solve each logarithmic equation. 17. log 4 (3 + ) = 3 18. 5 ln( ) = 0 19. log5 + log5 (4 1) = 1 0. log3 ( 5) + log3( + 3) = 1. log( + 4) log = log 4. log( 5 + 1) = log( + 3) + log 3. log( + 4) log = log(5 + 1) 4. ln( 4) + ln( + 1) = ln( 8) 5. How long will it take $000 to grow to $10,000 at 5% if compounded (a) continuously and () compounded quarterly. 6. What interest rate is needed for $1000 to doule after 10 years if compounded continuously? 0.0183t 7. The formula A =.9e models the population of Teas, A, in millions, t years after 005. (a) What was the population in Teas in 005? () Estimate the population in 01? (c) When will the population reach 7 million? 8. You have $300 to invest. What interest rate is needed for the investment to grow to $3000 in two years if the investment is compounded quarterly? 9. What interest is needed for an investment to triple in four years if it is compounded (a) semiannually and () continuously? 30. How much money would you need to invest now in order to have $15,000 saved in two years if the principal is invested at an interest rate of 6.% compounded (a) quarterly and () continuously? 31. A certain type of radioactive iodine has a half-life of 8 days. kt (a) Find an eponential decay model, A = A0e, for this type of iodine. Round the k value in your formula to si decimal places. () Use your model from part (a) to determine how long it will take for a sample of this type of radioactive iodine to decay to 10% of its original amount. Round your final answer to the nearest whole day. Page 10 (Section 4.6)
3. The population in Tanzania in 1987 was aout 4.3 million, with an annual growth rate of 3.5%. If the population is assumed to change continuously at this rate. (a) estimate the population in 008. () in how many years after 1987 will the population e 30,000,000? 33. The half-life of aspirin in your loodstream is 1 hours. How long will it take for the aspirin to decay to 70% of the original dosage? 34. The growth model A (a) What is Meico s growth rate? 0.01t = 107.4e descries Meico s population, A, in millions t years after 006. () How long will it take Meico to doule its population? 4.6 Homework Answers: 1. 7. 1 3. 3 4. 0 5. {.05} 6. {.90} 7. {.06} 8. {6.06} 6 9. {3.58} 10. {-.80} 11. {.8} 1. {.99} 13. {0, ln 3} 14. {0, ln } 15. {ln 3} 16. ln3 17. 6 3 18. 4 e 19. 5 4 0. {6} 1. 4 3. {5} 3. 9 4. φ 5. (a) 3.19 years; () 3.39 years 6. 6.93% 7. (a).9 million; () 6.03 million; (c) 014 8. 13.5% 9. (a) 9.44%; () 7.47% 30. (a) $13,63.31; () $13,50.70 31. (a) A = A e 0 0.086643t () 7 days 3. (a) 50.68 million; () 6.0 years 33. 6. hours 34. (a) 1.%; () aout 58 years Page 11 (Section 4.6)