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Splash Screen. Five-Minute Check (over Lesson 7–5) CCSS Then/Now New Vocabulary Example 1: Find Common Logarithms Example 2: Real-World Example: Solve Logarithmic Equations Example 3: Solve Exponential Equations Using Logarithms Example 4: Solve Exponential Inequalities Using Logarithms

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  1. Splash Screen

  2. Five-Minute Check (over Lesson 7–5) CCSS Then/Now New Vocabulary Example 1: Find Common Logarithms Example 2: Real-World Example: Solve Logarithmic Equations Example 3: Solve Exponential Equations Using Logarithms Example 4: Solve Exponential Inequalities Using Logarithms Key Concept: Change of Base Formula Example 5: Change of Base Formula Lesson Menu

  3. Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to approximate the value of log3 32. A. 1.9864 B. 2.3885 C. 3.1547 D. 4 5-Minute Check 1

  4. Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to approximate the value of log3 32. A. 1.9864 B. 2.3885 C. 3.1547 D. 4 5-Minute Check 1

  5. Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to approximate the value of log3 . 1 __ 2 • A. –0.6309 • B. 0.1577 • C. 0.3155 • 0.4732 5-Minute Check 2

  6. Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to approximate the value of log3 . 1 __ 2 • A. –0.6309 • B. 0.1577 • C. 0.3155 • 0.4732 5-Minute Check 2

  7. Solve log5 6 + 3 log5x = log5 48. A. 1 B. 2 C. 3 D. 4 5-Minute Check 3

  8. Solve log5 6 + 3 log5x = log5 48. A. 1 B. 2 C. 3 D. 4 5-Minute Check 3

  9. Solve log2 (n + 4) + log2n = 5. A. 10 B. 8 C. 6 D. 4 5-Minute Check 4

  10. Solve log2 (n + 4) + log2n = 5. A. 10 B. 8 C. 6 D. 4 5-Minute Check 4

  11. 1 __ Solve log6 16 – 2 log6 4 = log6 (x + 1) + log6 . 4 A. 2 B. 3 C. 3.5 D. 4 5-Minute Check 5

  12. 1 __ Solve log6 16 – 2 log6 4 = log6 (x + 1) + log6 . 4 A. 2 B. 3 C. 3.5 D. 4 5-Minute Check 5

  13. A.log8m5 = 5 log8m B.loga6 – loga 3 = loga 2 C. D.logb2x = logb 2 + logbx Which of the following equations is false? 5-Minute Check 6

  14. A.log8m5 = 5 log8m B.loga6 – loga 3 = loga 2 C. D.logb2x = logb 2 + logbx Which of the following equations is false? 5-Minute Check 6

  15. Content Standards A.CED.1 Create equations and inequalities in one variable and use them to solve problems. Mathematical Practices 4 Model with mathematics. CCSS

  16. You simplified expressions and solved equations using properties of logarithms. • Solve exponential equations and inequalities using common logarithms. • Evaluate logarithmic expressions using the Change of Base Formula. Then/Now

  17. common logarithm • Change of Base Formula Vocabulary

  18. ENTER LOG Keystrokes: 6 .7781512504 Find Common Logarithms A. Use a calculator to evaluate log 6 to the nearest ten-thousandth. Answer: Example 1

  19. ENTER LOG Keystrokes: 6 .7781512504 Find Common Logarithms A. Use a calculator to evaluate log 6 to the nearest ten-thousandth. Answer: about 0.7782 Example 1

  20. ENTER LOG Keystrokes: .35 –.4559319556 Find Common Logarithms B. Use a calculator to evaluate log 0.35 to the nearest ten-thousandth. Answer: Example 1

  21. ENTER LOG Keystrokes: .35 –.4559319556 Find Common Logarithms B. Use a calculator to evaluate log 0.35 to the nearest ten-thousandth. Answer: about –0.4559 Example 1

  22. A. Which value is approximately equivalent to log 5? A. 0.3010 B. 0.6990 C. 5.0000 D. 100,000.0000 Example 1

  23. A. Which value is approximately equivalent to log 5? A. 0.3010 B. 0.6990 C. 5.0000 D. 100,000.0000 Example 1

  24. B. Which value is approximately equivalent to log 0.62? A. –0.2076 B. 0.6200 C. 1.2076 D. 4.1687 Example 1

  25. B. Which value is approximately equivalent to log 0.62? A. –0.2076 B. 0.6200 C. 1.2076 D. 4.1687 Example 1

  26. JET ENGINES The loudness L, in decibels, of a sound is where I is the intensity of the sound and m is the minimum intensity of sound detectable by the human ear. The sound of a jet engine can reach a loudness of 125 decibels. How many times the minimum intensity of audible sound is this, if m is defined to be 1? Original equation Solve Logarithmic Equations Example 2

  27. Replace L with 125 and m with 1. Divide each side by 10 and simplify. Exponential form I ≈ 3.162 × 1012 Use a calculator. Solve Logarithmic Equations Answer: Example 2

  28. Replace L with 125 and m with 1. Divide each side by 10 and simplify. Exponential form I ≈ 3.162 × 1012 Use a calculator. Solve Logarithmic Equations Answer: The sound of a jet engine is approximately 3 × 1012 or 3 trillion times the minimum intensity of sound detectable by the human ear. Example 2

  29. DEMOLITION The loudness L, in decibels, of asound is where I is the intensity of the sound and m is the minimum intensity of sound detectable by the human ear. Refer to Example 2. The sound of the demolition of an old building can reach a loudness of 92 decibels. How many times the minimum intensity of audible sound is this, if m is defined to be 1? A. 1,585,000,000 times the minimum intensity B. 1,629,000,000 times the minimum intensity C. 1,912,000,000 times the minimum intensity D. 2,788,000,000 times the minimum intensity Example 2

  30. DEMOLITION The loudness L, in decibels, of asound is where I is the intensity of the sound and m is the minimum intensity of sound detectable by the human ear. Refer to Example 2. The sound of the demolition of an old building can reach a loudness of 92 decibels. How many times the minimum intensity of audible sound is this, if m is defined to be 1? A. 1,585,000,000 times the minimum intensity B. 1,629,000,000 times the minimum intensity C. 1,912,000,000 times the minimum intensity D. 2,788,000,000 times the minimum intensity Example 2

  31. Divide each side by log 5. Solve Exponential Equations Using Logarithms Solve 5x = 62. Round to the nearest ten-thousandth. 5x = 62 Original equation log 5x = log 62 Property of Equality for Logarithms x log 5 = log 62 Power Property of Logarithms x≈ 2.5643 Use a calculator. Answer: Example 3

  32. Divide each side by log 5. Solve Exponential Equations Using Logarithms Solve 5x = 62. Round to the nearest ten-thousandth. 5x = 62 Original equation log 5x = log 62 Property of Equality for Logarithms x log 5 = log 62 Power Property of Logarithms x≈ 2.5643 Use a calculator. Answer: about 2.5643 Example 3

  33. Solve Exponential Equations Using Logarithms Check You can check this answer by using a calculator or by using estimation. Since 52 = 25 and 53 = 125, the value of x is between 2 and 3. Thus, 2.5643 is a reasonable solution.  Example 3

  34. What is the solution to the equation 3x = 17? A.x = 0.3878 B.x = 2.5713 C.x = 2.5789 D.x = 5.6667 Example 3

  35. What is the solution to the equation 3x = 17? A.x = 0.3878 B.x = 2.5713 C.x = 2.5789 D.x = 5.6667 Example 3

  36. Solve Exponential Inequalities Using Logarithms Solve 37x > 25x – 3. Round to the nearest ten-thousandth. 37x > 25x – 3 Original inequality log 37x > log 25x – 3 Property of Inequality for Logarithmic Functions 7x log 3 > (5x – 3) log 2 Power Property of Logarithms Example 4

  37. Solve Exponential Inequalities Using Logarithms 7x log 3 > 5x log 2 – 3 log 2 Distributive Property 7x log 3 – 5x log 2 > – 3 log 2 Subtract 5x log 2 from each side. x(7 log 3 – 5 log 2) > –3 log 2 Distributive Property Example 4

  38. Divide each side by 7 log 3 – 5 log 2. Use a calculator. Solve Exponential Inequalities Using Logarithms x > –0.4922 Simplify. Example 4

  39. ? 37(0) > 25(0) – 3 Replace x with 0. ? 30 > 2–3 Simplify. Negative Exponent Property  Solve Exponential Inequalities Using Logarithms Check: Test x = 0. 37x > 25x – 3 Original inequality Answer: Example 4

  40. ? 37(0) > 25(0) – 3 Replace x with 0. ? 30 > 2–3 Simplify. Negative Exponent Property  Solve Exponential Inequalities Using Logarithms Check: Test x = 0. 37x > 25x – 3 Original inequality Answer: The solution set is {x | x > –0.4922}. Example 4

  41. What is the solution to 53x < 10x – 2? A. {x | x > –1.8233} B. {x | x < 0.9538} C. {x | x > –0.9538} D. {x | x < –1.8233} Example 4

  42. What is the solution to 53x < 10x – 2? A. {x | x > –1.8233} B. {x | x < 0.9538} C. {x | x > –0.9538} D. {x | x < –1.8233} Example 4

  43. Concept

  44. Change of Base Formula Use a calculator. Change of Base Formula Express log5 140 in terms of common logarithms. Then round to the nearest ten-thousandth. Answer: Example 5

  45. Change of Base Formula Use a calculator. Change of Base Formula Express log5 140 in terms of common logarithms. Then round to the nearest ten-thousandth. Answer: The value of log5 140 is approximately 3.0704. Example 5

  46. A. B. C. D. What is log5 16 expressed in terms of common logarithms? Example 5

  47. A. B. C. D. What is log5 16 expressed in terms of common logarithms? Example 5

  48. End of the Lesson

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