Chapter 11 - Lesson 11.1   Skills Practice Name Date All...

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Unformatted text preview: Lesson 11.1   Skills Practice Name Date All the Pieces of the Puzzle Exponential and Logarithmic Forms Vocabulary 1. Give an example of a logarithmic equation. Problem Set Arrange the given terms to create a true exponential equation and a true logarithmic equation. 2. 3, 11, 1331 1. 25, 5, 2 Exponential equation: 5​ ​ 5 25 2 Logarithmic equation: ​log​5​25 5 2 11 4. 1296, 4, 6 5. 23, ____ ​  1   ​, 8 512 1  ​, 4, 16 6. ​ __ 2 7. 17, ____ ​  1   ​, 22 289 8. __ ​ 1 ​ , 9, 729 3 © Carnegie Learning 3. 2, 256, 8 Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 861   861 28/11/13 4:31 PM Lesson 11.1   Skills Practice page 2 Solve for the unknown. 9. ​log​7​343 5 n ​log​7​343 5 n 10. ​log​__1​64 5 n ​   ​  4 ​ ​n​ 5 343 7 ​ ​n​ 5 ​73​ ​ 7 n 5 3 (  ) 11 ​log​n​1024 5 5 12. ​log​n​ ​____ ​  1   ​  ​ 5 24 625 3   )​5 __   ​4 8 ​ ​   ​  13. ​log​n​  ​( 4 14. log n 5 6 15. ​log​4​16 5 n 16. ​log​81​​__ ​ 1 ​   ​5 n 9 862  (  ) © Carnegie Learning 11   Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 862 28/11/13 4:31 PM Lesson 11.1   Skills Practice Name page 3 Date Estimate the logarithm to the tenths place. Explain how you determined your answer. 17. ​log​2​48 The logarithm ​log​2​48 is approximately equal to 5.5. ​log​2​32 , ​log​2​48 , ​log​2​64   5   ,  n  ,  6 Because 48 is halfway between 32 and 64, the approximation should be halfway between 5 and 6. 18. ​log​3​2 11 ( ____ ) © Carnegie Learning 19. ​log​8​​​  1   ​ ​ 495 20. ​log​6​53 Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 863   863 28/11/13 4:31 PM Lesson 11.1   Skills Practice page 4 21. ​log​9​6000 (  ) 22. ​log​7​​___ ​ 1  ​  ​ 10 (  ) 23. ​log​__1​​___ ​  1  ​  ​ ​   ​  24 2 24. ​log​__4​0.85 ​   ​  3 864  © Carnegie Learning 11   Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 864 28/11/13 4:31 PM Lesson 11.1   Skills Practice Name page 5 Date Determine the appropriate base of each logarithm. Explain your reasoning. 25. ​log​b​29 5 2.1 ​log​5​25 , ​log​b​29 , ​log​5​125    2  ,  2.1  ,  3 Because the value of the exponent is 2.1, that means that the argument 29 should be close to its lower limit. When the base is 5, 29 is very close to the lower limit of 25, whereas when the base is 4, 29 is not very close to the lower limit of 16. © Carnegie Learning 26. ​log​b​35 5 1.9 11 (  ) 27. ​log​b​​__ ​ 1 ​   ​5 21.9 7 Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 865   865 28/11/13 4:31 PM Lesson 11.1   Skills Practice page 6 28. ​log​b​80 5 3.9 29. ​log​b​6 5 0.9 11 © Carnegie Learning 30. ​log​b​66 5 3.1 866    Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 866 28/11/13 4:31 PM Lesson 11.2   Skills Practice Name Date Mad Props Properties of Logarithms Problem Set Rewrite each logarithmic expression in expanded form using the properties of logarithms. 1. ​log​3​(​ 5x )​ ​log​3​(5x) 5 ​log​3​5 1 ​log​3​x (  ) 3. ​log​7​​( ​n4​ ​ )​ 4. log ​__ ​ x  ​ ​ 7 5. ​log​2​(​ mn )​ 6. log (​ ​pq​ ​ )​ 7. ln (​ ​x2​ ​ )​ 8. ln ​__ ​ c  ​ ​ 3 9. ​log​3​​( ​7x​2​ )​ 10. ln (​ ​2x​3​​y​2​ )​ (  ) xy ​   ​  ​ 11. log ​___ 5 © Carnegie Learning (  ) 2. ​log​5​​__ ​ a ​​ b (  ) 11 (  ) (  ) 4 12. ​log​7​​___ ​ ​3yx​  ​  ​ ​ (  ) ​  x   ​​ 13. ln ​___ 7y 2 14. ​log​5​​___ ​ 7​x3​ ​ ​ ​ ​y​ ​ 15. log ​( xyz )​ 16. ln ​_______ ​  x 1 1  ​  ​   ​(y 1 3)​2​ (  ) Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 867   867 28/11/13 4:31 PM Lesson 11.2   Skills Practice page 2 Rewrite each logarithmic expression as a single logarithm. 17. log x 2 2 log y log x 2 2 log y 5 log ​ __ ( ​ ​yx​  ​ ​ )​ 18. 3 ​log​4​x 1 ​log​4​y 2 ​log​4​z 2 19. 6 ​log​2​x 2 2 ​log​2​x 20. log 3 1 2 log 7 2 log 6 21. log x 1 3 log y 2 __ ​ 1 ​ log z 2 22. 7 ​log​3​x 2 (2 ​log​3​x 1 5 ​log​3​y) 23. 2 ln (2x 1 3) 2 4 ln (y 2 2) 24. ln (x 2 7) 2 2(ln x 1 ln y) 11 Suppose w 5 log​b​2, x 5 log​b​3, y 5 log​b​7, and z 5 log​b​11. Write an algebraic expression for each logarithmic expression. 26. ​log​b​98 25. ​log​b​33 ​log​b​33 5 ​log​b​(3 ? 11) 5 ​log​b​3 1 ​log​b​11 (  ) 868  (  ) ​ 2 ​   ​ 27. ​log​b​​__ 3 28. ​log​b​​__ ​ 7  ​ ​ 8 29. ​log​b​1.5 30. ​log​b​2.75 © Carnegie Learning 5 x 1 z   Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 868 28/11/13 4:31 PM Lesson 11.3   Skills Practice Name Date What’s Your Strategy? Solving Exponential Equations Vocabulary 1. Define the Change of Base Formula and explain how it is used. Problem Set Solve each exponential equation by using the Change of Base Formula. 1. ​6​x25​5 24 2. ​12​x14​5 65 ​6x25 ​ ​ 5 24 x 2 5 5 ​log​6​24 11 ______  log 24  ​  x 2 5 5 ​  log 6 x 2 5 ¯ 1.774 x ¯ 6.774 4. 8 ​ 5x ​ ​5 71 © Carnegie Learning 3. 7 ​ 3x ​ ​5 15 Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 869   869 28/11/13 4:31 PM Lesson 11.3   Skills Practice 5. 4 ​ x13 ​ ​2 7 5 32 (  ) page 2 6. ​11​x24​1 8 5 59 (  ) 7. 4​​__ ​ 2 ​   ​​ ​5 248 3 8. 9​​__ ​ 3 ​   ​​ ​5 999 5 9. ​2(5)​2x11​1 4 5 18 10. ​28(2)​x29​25 5 277 3x 2x © Carnegie Learning 11 870    Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 870 28/11/13 4:31 PM Lesson 11.3   Skills Practice page 3 Name Date Solve each exponential equation using properties of logarithms. 12. ​15​x12​5 60 11. 7 ​ x22 ​ ​5 28 ​ x22 7 ​ ​ 5 28 (x 2 2) log 7 5 log 28 ______  log 28  ​  x 2 2 5 ​  log 7 x 2 2 ¯ 1.712 x ¯ 3.712 13. 5 ​ ​4x​5 32 14. 9 ​ ​3x​5 124 11 16. ​14​x26​1 3 5 73 © Carnegie Learning 15. ​3​x17​2 5 5 63 Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 871   871 28/11/13 4:31 PM Lesson 11.3   Skills Practice (  ) 11 page 4 (  ) 2x ​ 4 ​   ​​ ​5 348 17. ​6​__ 7 4x 18. ​8 ​__ ​ 5 ​   ​​ ​5 448 8 19. ​3(4)​3x26​1 2 5 35 20. ​27(3)​x16​2 8 5 2162 © Carnegie Learning 872    Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 872 28/11/13 4:31 PM Lesson 11.3   Skills Practice page 5 Name Date Solve each exponential equation. Explain why you chose the method that you used. 21. ​11​x24​5 343 ​11​x24​ 5 343 (x 2 4) log 11 5 log 343 log 343  ​  x 2 4 5 ​  log 11 x 2 4 ¯ 2.435 _______  x ¯ 6.435 I took the log of both sides, because 343 cannot be written as a power of 11. 22. 5 ​ ​2x+1​5 3125 11 (  ) © Carnegie Learning 5x 23. ​16​__ ​ 8 ​   ​​ ​5 752 9 Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 873   873 28/11/13 4:31 PM Lesson 11.3   Skills Practice page 6 24. ​23​5x​5 736 (  ) 4x 25. ​​___ ​ 7  ​  ​​ ​5 49 12 11 © Carnegie Learning 26. ​14​5x212​2 8 5 2736 874    Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 874 28/11/13 4:31 PM Lesson 11.4   Skills Practice Name Date Logging On Solving Logarithmic Equations Problem Set Solve each logarithmic equation. Check your answer(s). 1. ​log​2​(​x2​ ​2 x) 5 1 2. ​log​15​(​x2​ ​2 2x) 5 1 ​log​2​(​x2​ ​2 x) 5 1 ​ 1​ ​ 5 ​x2​ ​2 x 2 0 5 ​x2​ ​2 x 2 2 0 5 (x 1 1)(x 2 2) x 5 21, 2 Check: ​log​2​​( ​(21)​2​2 (21) )​ 0 1 11 ​log​2​(1 1 1) 0 1 ​log​2​2 5 1 ​log​2​(​22​ ​2 2) 0 1 ​log​2​(4 2 2) 0 1 © Carnegie Learning ​log​2​2 5 1 Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 875   875 28/11/13 4:31 PM Lesson 11.4   Skills Practice page 2 3. ​log​6​(​x2​ ​1 5x) 5 2 4. ​log​2​(​x2​ ​1 6x) 5 4 5. ​log​4​(​x2​ ​2 12x) 5 3 6. ​log​10​(​x2​ ​1 15x) 5 2 © Carnegie Learning 11 876    Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 876 28/11/13 4:31 PM Lesson 11.4   Skills Practice page 3 Name 7. ​log​3​(​3x​2​118x) 5 4 Date 8. ​log​4​(​2x​2​2 28x) 5 3 © Carnegie Learning 11 Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 877   877 28/11/13 4:31 PM Lesson 11.4   Skills Practice page 4 Use the properties of logarithms to solve each logarithmic equation. Check your answer(s). 9. 2 ​log​3​x 2 ​log​3​8 5 ​log​3​(x 2 2) 10. ​log​4​(x 1 3) 1 ​log​4​x 5 1 2 ​log​3​x 2 ​log​3​8 5 ​log​3​(x 2 2) ​log​3​​x​2​2 ​log​3​8 5 ​log​3​(x 2 2) 2 ​log​3​​ ​ ​x​ ​ ​ ​ 5 ​log​3​(x 2 2) 8 ( __ ) __​ ​x​ ​​   5 x 2 2 8 2 x​ 2​ ​ 5 8x 2 16 ​x2​ ​2 8x 1 16 5 0 ​(x 2 4)​2​ 5 0 x 5 4 Check: 2 ​log​3​4 2 ​log​3​8 0 ​log​3​(4 2 2) ​log​3​16 2 ​log​3​8 0 ​log​3​2 ​log​3​​ ​ 16 ​  ​ 0 ​log​3​2 8 ​log​3​2 5 ​log​3​2 ( ___ ) © Carnegie Learning 11 878    Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 878 28/11/13 4:31 PM Lesson 11.4   Skills Practice page 5 Name 11. log (2​x2​ ​1 3) 1 log 2 5 log 10x Date 12. ​log​2​x 1 ​log​2​(x 2 6) 5 4 © Carnegie Learning 11 Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 879   879 28/11/13 4:31 PM Lesson 11.4   Skills Practice 13. 2 ​log​5​x 2 ​log​5​4 5 log​5​(8 2 x) page 6 14. ​log​2​3 1 ​log​2​(​3x​2​1 4) 5 log​2​​( 39x )​ © Carnegie Learning 11 880    Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 880 28/11/13 4:31 PM Lesson 11.4   Skills Practice page 7 Name Date (  ) 15. ln ​​x2​ ​1 ___ ​ 15 ​  ​1 ln 2 5 ln (11x) 2 (  ) (  ) 16. ​log​4​​__ ​ 1 ​  ​x2​ ​2 6 ​2 ​log​4​​__ ​ 1 ​   ​5 log​4​x 5 5 © Carnegie Learning 11 Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 881   881 28/11/13 4:31 PM © Carnegie Learning 11 882    Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 882 28/11/13 4:31 PM Lesson 11.5   Skills Practice Name Date So When Will I Use This? Applications of Exponential and Logarithmic Equations Problem Set The amount of a radioactive isotope remaining can be modeled using the formula A 5 A ​ ​0​​e​2kt​, where t represents the time in years, A represents the amount of the isotope remaining in grams after t years, ​A0​ ​represents the original amount of the isotope in grams, and k is the decay constant. Use this formula to solve each problem. 1. Strontium-90 is a radioactive isotope with a half-life of about 29 years. Calculate the decay constant for Strontium-90. Then find the amount of 100 grams of Strontium-90 remaining after 120 years. A 5 ​A0​ ​​e​2kt​ ​ 1 ​ ​A​0​ 5 ​A0​ ​​e​2k(29)​ 2 ​ ​ ​ 1  ​ 5 ​e229k 2 ln ​​ 1 ​   ​ 5 229k 2 __  __  ( __ ) 11 k ¯ 0.0239 The decay constant for Strontium-90 is about 0.0239. A 5 100​e​20.0239(120)​ A ¯ 5.681 After 120 years, there would be about 5.681 grams remaining. © Carnegie Learning 2. Radium-226 is a radioactive isotope with a half-life of about 1622 years. Calculate the decay constant for Radium-226. Then find the amount of 20 grams of Radium-226 remaining after 500 years. Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 883   883 28/11/13 4:31 PM Lesson 11.5   Skills Practice page 2 3. Carbon-14 is a radioactive isotope with a half-life of about 5730 years. Calculate the decay constant for Carbon-14. Then find the amount of 6 grams of Carbon-14 that will remain after 22,000 years. 4. Cesium-137 is a radioactive isotope with a half-life of about 30 years. Calculate the decay constant for Cesium-137. Then calculate the percentage of a Cesium-137 sample remaining after 100 years. © Carnegie Learning 11 884    Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 884 28/11/13 4:31 PM Lesson 11.5   Skills Practice Name page 3 Date 5. Uranium-232 is a radioactive isotope with a half-life of about 69 years. Calculate the decay constant for Uranium-232. Then calculate the percentage of a Uranium-232 sample remaining after 200 years. 11 © Carnegie Learning 6. Rubidium-87 is a radioactive isotope with a half-life of about 4.7 3 ​10​7​years. Calculate the decay constant for Rubidium-87. Then calculate the percentage of a Rubidium-87 sample remaining after 1,000,000 years. Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 885   885 28/11/13 4:31 PM Lesson 11.5   Skills Practice page 4 Use the given exponential equation to answer each question. Show your work. 7. The number of students exposed to the measles at a school can be modeled by the equation S 5 10​e0.15t ​ ​, where S represents the number of students exposed after t days. How many students were exposed after eight days? S 5 10​e0.15t ​ ​ 5 10​e​(0.15 ? 8)​ 5 10​e​1.2​ ¯ 33.20116923 Approximately 33 students were exposed after eight days. 8. The minnow population in White Mountain Lake each year can be modeled by the equation M 5 700(​10​0.2t​) , where M represents the minnow population t years from now. What will the minnow population be in 15 years? 9. Aiden invested $600 in a savings account with continuous compound interest. The equation V 5 600​e0.05t ​ ​can be used to predict the value, V, of Aiden’s account after t years. What would the value of Aiden’s account be after five years? 886  © Carnegie Learning 11   Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 886 28/11/13 4:31 PM Lesson 11.5   Skills Practice Name page 5 Date 10. The rabbit population on Hare Island can be modeled by the equation R 5 60​e​0.09t​, where R represents the rabbit population t years from now. How many years from now will the rabbit population of Hare Island be 177 rabbits? 11 © Carnegie Learning 11. A disease is destroying the elm tree population in the Dutch Forest. The equation N 5 16(​10​0.15t​) can be used to predict the number of elm trees, N, killed by the disease t years from now. In how many years from now will 406 elm trees have been killed by the disease? Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 887   887 28/11/13 4:31 PM Lesson 11.5   Skills Practice page 6 12. Manuel invested money in a savings account with continuous compound interest. The equation V 5 10,000​e0.03t ​ ​can be used to determine the value, V, of the account after t years. In how many years will the value of the account be $12,000? (  ) 11 Use the formula M 5 log ​__ ​ I  ​  ​, where M is the magnitude of an earthquake on the Richter scale, ​I0​ ​ ​I0​ ​ represents the intensity of a zero-level earthquake the same distance from the epicenter, and I is the number of times more intense an earthquake is than a zero-level earthquake, to solve each problem. A zero-level earthquake has a seismographic reading of 0.001 millimeter at a distance of 100 kilometers from the center. 13. An earthquake southwest of Chattanooga, Tennessee in 2003 had a seismographic reading of 79.43 millimeters registered 100 kilometers from the center. What was the magnitude of the Tennessee earthquake of 2003 on the Richter scale? ( __ ) M 5 log ​ ​ I  ​  ​ ​I0​ ​ ( ______ ) M 5 log ​ ​ 79.43 ​   ​ 0.001 M ¯ 4.9 14. An earthquake in Illinois in 2008 had a seismographic reading of 158.5 millimeters registered 100 kilometers from the center. What was the magnitude of the Illinois earthquake of 2008 on the Richter scale? 888  © Carnegie Learning The Tennessee earthquake of 2003 measured 4.9 on the Richter scale.   Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 888 28/11/13 4:31 PM Lesson 11.5   Skills Practice Name page 7 Date 15. An earthquake off the northern coast of California in 2005 had a seismographic reading of 15,849 millimeters registered 100 kilometers from the center. What was the magnitude of the California earthquake in 2005 on the Richter scale? 16. The devastating earthquake in Haiti in 2010 had a magnitude of 7.0 on the Richter scale. What was its seismographic reading in millimeters 100 kilometers from the center? © Carnegie Learning 11 17. Calculate the value of the seismographic reading for an earthquake of magnitude 6.4 on the Richter scale. Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 889   889 28/11/13 4:31 PM Lesson 11.5   Skills Practice page 8 18. Calculate the value of the seismographic reading for an earthquake of magnitude 8.1 on the Richter scale. Use the given formula to solve each problem. 19. The formula for the population of a species is n 5 k log (A), where n represents the population of a species, A is the area of the region in which the species lives, and k is a constant that is determined by field studies. Based on population samples, an area that is 1000 square miles contains 360 wolves. Calculate the value of k. Then use the formula to find the number of wolves remaining in 15 years if only 300 square miles of this area is still inhabitable. 11 n 5 k log (A) 360 5 k log 1000 k 5 120 The value of k is 120. n 5 120 log (A) 5 120 log 300 ¯ 297 20. The formula for the population of a species is n 5 k log (A), where n represents the population of a species, A is the area of the region in which the species lives, and k is a constant that is determined by field studies. Based on population samples, a rainforest that is 100 square miles contains 342 monkeys. Calculate the value of k. Then use the formula to find the number of monkeys remaining in 5 years if only 40 square miles of the rainforest survives due to the current level of deforestation. 890  © Carnegie Learning In 15 years, there will be approximately 297 wolves remaining in the area.   Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 890 28/11/13 4:31 PM Lesson 11.5   Skills Practice Name page 9 Date 21. The formula y 5 a 1 b ln t, where t represents the time in hours, y represents the amount of fresh water produced in t hours, a represents the amount of fresh water produced in one hour, and b is the rate of production, models the amount of fresh water produced from salt water during a desalinization process. In one desalination plant, 15.26 cubic yards of fresh water can be produced in one hour with a rate of production of 31.2. How much fresh water can be produced after 8 hours? 11 © Carnegie Learning 22. The formula y 5 a 1 b ln t, where t represents the time in hours, y represents the amount of fresh water produced in t hours, a represents the amount of fresh water produced in one hour, and b is the rate of production, models the amount of fresh water produced from salt water during a desalinization process. At a desalination plant, 18.65 cubic yards of fresh water can be produced in one hour with a rate of production of 34.5. How long will it take for the plant to produce 250 cubic yards of fresh water? Chapter 11  Skills Practice  504371_Skills_CH11_861-892.indd 891   891 28/11/13 4:31 PM Lesson 11.5   Skills Practice page 10 23. The relationship between the age of an item in years and its value is given by the equation ​ V ​​ log ​__ C __________ t5    ​  , where t represents the age of the item in years, V represents the value of the item log (1 2 r) after t years, C represents the original value of the item, and r represents the yearly rate of appreciation expressed as a decimal. A luxury car was originally purchased for $110,250 and is currently valued at $65,200. The average rate of depreciate for this car is 10.3% per year. How old is the car to the nearest tenth of a year? (  ) 11 24. The relationship between the age of an item in years and its value is given by the equation ​ V ​​ log ​__ C __________ t5    ​  , where t represents the age of the item in years, V represents the value of the item after log (1 2 r) t years, C represents the original value of the item, and r represents the yearly rate of appreciation expressed as a decimal. A 4-year old car was originally purchased for $35,210. Its current value is $16,394. What is this car’s annual rate of depreciation to the nearest tenth? © Carnegie Learning (  ) 892    Chapter 11  Skills Practice 504371_Skills_CH11_861-892.indd 892 28/11/13 4:31 PM ...
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