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1. Exponential and Logarithmic Equations and Inequalities 4-5 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

2. Review assignment

3. 3 2 Warm Up Solve. 1.log16x = 2. log10,000 =x 64 4

4. Objectives Solve exponential and logarithmic equations and equalities. Solve problems involving exponential and logarithmic equations.

5. Vocabulary exponential equation logarithmic equation

6. An exponential equation is an equation containing one or more expressions that have a variable as an exponent. To solve exponential equations: • Try writing them so that the bases are all the same. • Take the logarithm of both sides.

7. Helpful Hint When you use a rounded number in a check, the result will not be exact, but it should be reasonable.

8. Example 1A: Solving Exponential Equations Solve and check. 98 – x = 27x – 3 (32)8 – x = (33)x – 3 Rewrite each side with the same base; 9 and 27 are powers of 3. 316– 2x = 33x – 9 To raise a power to a power, multiply exponents. 16 – 2x = 3x – 9 Bases are the same, so the exponents must be equal. Solve for x. x = 5

9. Example 1A Continued Check 98 – x = 27x – 3 98 – 5275 – 3 93 272  729 729 The solution is x = 5.

10. log5 log5 x–1 = log4 log4 x = 1 + ≈ 2.161 Example 1B: Solving Exponential Equations Solve and check. 4x – 1= 5 5 is not a power of 4, so take the log of both sides. log 4x – 1 = log 5 (x– 1)log 4 = log 5 Apply the Power Property of Logarithms. Divide both sides by log 4. CheckUse a calculator. The solution is x ≈ 2.161.

11. You Try! Example 1a Solve and check. 32x = 27 Rewrite each side with the same base; 3 and 27 are powers of 3. (3)2x = (3)3 To raise a power to a power, multiply exponents. 32x = 33 2x = 3 Bases are the same, so the exponents must be equal. x = 1.5 Solve for x.

12. You Try! Example 1a Continued Check 32x = 27 32(1.5)27 33 27  27 27 The solution is x = 1.5.

13. log21 log21 –x = log7 log7 x = – ≈ –1.565 You Try! Example 1b Solve and check. 7–x = 21 21 is not a power of 7, so take the log of both sides. log 7–x = log 21 Apply the Power Property of Logarithms. (–x)log 7 = log 21 Divide both sides by log 7.

14. You Try! Example 1b Continued CheckUse a calculator. The solution is x ≈ –1.565.

15. log15 3x = log2 You Try! Example 1c Solve and check. 23x = 15 15 is not a power of 2, so take the log of both sides. log23x = log15 Apply the Power Property of Logarithms. (3x)log 2 = log15 Divide both sides by log 2, then divide both sides by 3. x ≈ 1.302

16. You Try! Example 1c Continued CheckUse a calculator. The solution is x ≈ 1.302.

17. Example 2: Biology Application Suppose a bacteria culture doubles in size every hour. How many hours will it take for the number of bacteria to exceed 1,000,000? At hour 0, there is one bacterium, or 20 bacteria. At hour one, there are two bacteria, or 21 bacteria, and so on. So, at hour n there will be 2n bacteria. Write 1,000,000 in scientific annotation. Solve 2n > 106 log2 2n > log2 106 Take the log2 of both sides.

18. Example 2 Continued log2 2n > log2 106 n> log2 106 Log2 2n is n. Evaluate by using change of base and a calculator. > n > ≈ 19.94 Round up to the next whole number. It will take about 20 hours for the number of bacteria to exceed 1,000,000.

19. Example 2 Continued Check In 20 hours, there will be 220 bacteria. 220 = 1,048,576 bacteria.

20. You Try! Example 2 You receive one penny on the first day, and then triple that (3 cents) on the second day, and so on for a month. On what day would you receive at least a million dollars. \$1,000,000 is 100,000,000 cents. On day 1, you would receive 1 cent or 30 cents. On day 2, you would receive 3 cents or 31 cents, and so on. So, on day n you would receive 3n–1cents. Write 100,000,000 in scientific annotation. Solve 3n – 1 > 1 x 108 log3 3n –1 > log3 108 Take the log3 of both sides.

21. 8 8 log 3 log3 n – 1 > n > + 1 You Try! Example 2 Continued log3 3n –1 > log3 108 n – 1> log3 108 log3 3n-1 is n-1. log Evaluate by using change of base and a calculator. log Round up to the next whole number. n > ≈ 17.8 Beginning on day 18, you would receive more than a million dollars.

22. You Try! Example 2 Check On day 18, you would receive 318 – 1 or over a million dollars. 317 = 129,140,163 cents or 1.29 million dollars.

23. A logarithmic equation is an equation with a logarithmic expression that contains a variable. You can solve logarithmic equations by using the properties of logarithms.

24. 1 7 12 6 x= Example 3A: Solving Logarithmic Equations Solve. log6(2x – 1) = –1 6–1= (2x –1) Put in exponential form. = 2x – 1 Simplify.

25. 100 x + 1 log4( ) = 1 Example 3B: Solving Logarithmic Equations Solve. log4100 – log4(x + 1) = 1 Write as a quotient. Write in exponential form. 4x+4=100 4x=96 x= 24

26. Example 3C: Solving Logarithmic Equations Solve. log5x 4 = 8 4log5x = 8 Power Property of Logarithms. log5x = 2 Divide both sides by 4 to isolate log5x. x = 52 Definition of a logarithm. x= 25

27. Example 3D: Solving Logarithmic Equations Solve. log12x+ log12(x + 1) = 1 Product Property of Logarithms. log12x(x + 1) = 1 121=x(x+1) Exponential form.

28. Example 3 Continued x2 + x – 12 = 0 Multiply and collect terms. (x – 3)(x+ 4) = 0 Factor. Set each of the factors equal to zero. x – 3 = 0 or x+ 4 = 0 x = 3 or x= –4 Solve. Check Check both solutions in the original equation. log12x+ log12(x +1) = 1 log12x+ log12(x +1) = 1 x log123+ log12(3 + 1) 1 log12( –4) + log12(–4 +1) 1 log123 + log124 1 log12( –4) is undefined. log1212 1 1 1  The solution is x = 3.

29. You Try! Example 3a Solve. 3 = log 8 + 3log x 3 = log 8 + 3log x 3 = log 8 + log x3 Power Property of Logarithms. 3 = log (8x3) Product Property of Logarithms. 103 = (8x3) Exponential Form 1000 = 8x3 125 = x3 5 = x

30. You Try! Example 3b Solve. 2log x– log 4 = 0 Condense. Exponential Form Only Positive 2 Not Negative 2 x= 2

31. Assignment: Solving Exponential and Log Equations Worksheet

32. Objective - To solve exponential and log equations. Rewrite each base as a power of 2. Distribute. Powers equal to each other.

33. Take the log3 of each side.

34. Take the log5 of each side.

35. If logb x = logb y if and only if x = y.

36. Rewrite in exponential form

37. Check!

38. Check Point 251.189 1000 14 39.121 -1, 2 about 1.082

39. Assignment Section 8.6 #32-42 even,46-58 even p 505-506

40. Assignment Section 8.6 #32-42 even,46-58 even p 505-506

41. Answers Section 8.6 #32-42 even,46-58 even p 505-506 46. 32. 48. 50. 34. 42. 52. 36. 54. 56. 38. 58. 40.

42. Caution Watch out for calculated solutions that are not solutions of the original equation.

43. Example 4A: Using Tables and Graphs to Solve Exponential and Logarithmic Equations and Inequalities Use a table and graph to solve 2x + 1 > 8192x. Use a graphing calculator. Enter 2^(x + 1)as Y1 and8192xas Y2. In the graph, find the x-value at the point of intersection. In the table, find the x-values where Y1 is greater than Y2. The solution set is {x | x > 16}.

44. Use a graphing calculator. Enter log(x + 70) as Y1 and 2log() as Y2. x x 3 3 Example 4B log(x + 70) = 2log() In the graph, find the x-value at the point of intersection. In the table, find the x-values where Y1 equals Y2. The solution is x = 30.

45. You Try! Example 4a Use a table and graph to solve 2x = 4x – 1. Use a graphing calculator. Enter 2xas Y1 and4(x – 1)as Y2. In the graph, find the x-value at the point of intersection. In the table, find the x-values where Y1 is equal to Y2. The solution is x = 2.

46. You Try! Example 4b Use a table and graph to solve 2x > 4x – 1. Use a graphing calculator. Enter 2xas Y1 and4(x – 1)as Y2. In the graph, find the x-value at the point of intersection. In the table, find the x-values where Y1 is greater than Y2. The solution is x < 2.

47. You Try! Example 4c Use a table and graph to solve log x2 = 6. Use a graphing calculator. Enter log(x2)as Y1 and6 as Y2. In the graph, find the x-value at the point of intersection. In the table, find the x-values where Y1 is equal to Y2. The solution is x = 1000.