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Splash Screen. Five-Minute Check (over Lesson 7–4) CCSS Then/Now Key Concept: Product Property of Logarithms Example 1: Use the Product Property Key Concept: Quotient Property of Logarithms Example 2: Real-World Example: Quotient Property Key Concept: Power Property of Logarithms

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  1. Splash Screen

  2. Five-Minute Check (over Lesson 7–4) CCSS Then/Now Key Concept: Product Property of Logarithms Example 1: Use the Product Property Key Concept: Quotient Property of Logarithms Example 2: Real-World Example: Quotient Property Key Concept: Power Property of Logarithms Example 3: Power Property of Logarithms Example 4: Solve Equations Using Properties of Logarithms Lesson Menu

  3. Solve log4 (x2 – 30) = log4x. A.x = 7 B.x = 6 C.x = 5 D.x = 4 5-Minute Check 1

  4. Solve log4 (x2 – 30) = log4x. A.x = 7 B.x = 6 C.x = 5 D.x = 4 5-Minute Check 1

  5. Solve log5 (x2 – 2x) = log5 (–5x + 10). A.x = 2 B.x = 1 C.x = –5 D.x = –10 5-Minute Check 2

  6. Solve log5 (x2 – 2x) = log5 (–5x + 10). A.x = 2 B.x = 1 C.x = –5 D.x = –10 5-Minute Check 2

  7. Solve log3x < 3. A. {x | 0 < x < 27} B. {x | 0 < x < 18} C. {x | 0 < x < 9} D. {x | 0 < x < 6} 5-Minute Check 3

  8. Solve log3x < 3. A. {x | 0 < x < 27} B. {x | 0 < x < 18} C. {x | 0 < x < 9} D. {x | 0 < x < 6} 5-Minute Check 3

  9. Solve log9 (4x + 6) > log9 (x + 12). A. {x | x < 3} B. {x | x > 3} C. {x | x < 2} D. {x | x > 2} 5-Minute Check 4

  10. Solve log9 (4x + 6) > log9 (x + 12). A. {x | x < 3} B. {x | x > 3} C. {x | x < 2} D. {x | x > 2} 5-Minute Check 4

  11. A.{x | –1 < x < 2} B. C. D.{x | 1 < x < 2} Solve log7 (x + 3) ≥ log7 (6x – 2). 5-Minute Check 5

  12. A.{x | –1 < x < 2} B. C. D.{x | 1 < x < 2} Solve log7 (x + 3) ≥ log7 (6x – 2). 5-Minute Check 5

  13. A.–1 B. – C. D.3 1 3 __ __ 2 4 Which of the following is a solution to the inequality log8 (x – 2) ≤ log8 (5x – 6)? 5-Minute Check 6

  14. A.–1 B. – C. D.3 1 3 __ __ 2 4 Which of the following is a solution to the inequality log8 (x – 2) ≤ log8 (5x – 6)? 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 8 Look for and express regularity in repeated reasoning. CCSS

  16. You evaluated logarithmic expressions and solved logarithmic equations. • Simplify and evaluate expressions using the properties of logarithms. • Solve logarithmic equations using the properties of logarithms. Then/Now

  17. Concept

  18. Use the Product Property Use log5 2 ≈ 0.4307 to approximate the value of log5 250. log5 250 = log5 (53 ● 2) Replace 250 with 53 ● 2. = log5 53 + log5 2 Product Property = 3 + log5 2 Inverse Property of Exponents and Logarithms ≈3 + 0.4307 or 3.4307 Replace log5 2 with 0.4307. Answer: Example 1

  19. Use the Product Property Use log5 2 ≈ 0.4307 to approximate the value of log5 250. log5 250 = log5 (53 ● 2) Replace 250 with 53 ● 2. = log5 53 + log5 2 Product Property = 3 + log5 2 Inverse Property of Exponents and Logarithms ≈3 + 0.4307 or 3.4307 Replace log5 2 with 0.4307. Answer: Thus, log5 250 is approximately 3.4307. Example 1

  20. Given log2 3 ≈ 1.5850, what is the approximate value of log2 96? A. –3.415 B. 3.415 C. 5.5850 D. 6.5850 Example 1

  21. Given log2 3 ≈ 1.5850, what is the approximate value of log2 96? A. –3.415 B. 3.415 C. 5.5850 D. 6.5850 Example 1

  22. Concept

  23. SCIENCEThe pH of a substance is defined as the concentration of hydrogen ions [H+] in moles. It is given by the formula pH = . Find the amount of hydrogen in a liter of acid rain that has a pH of 5.5. Quotient Property Example 2

  24. Original equation Substitute 5.5 for pH. Quotient Property log101 = 0 Quotient Property Understand The formula for finding pH and the pH of the rain is given. Plan Write the equation. Then, solve for [H+]. Solve Example 2

  25. H+ H+ H+ Quotient Property Simplify. Multiply each side by –1. Definition of logarithm Answer: Example 2

  26. H+ H+ H+ Quotient Property Simplify. Multiply each side by –1. Definition of logarithm Answer: There are 10–5.5, or about 0.0000032, mole of hydrogen in a liter of this rain. Example 2

  27. ? 5.5 = log101 – log1010–5.5 Quotient Property ? 5.5 = 0 – (–5.5)Simplify. ? Quotient Property Check pH = 5.5 H+ = 10–5.5 5.5 = 5.5  Example 2

  28. SCIENCE The pH of a substance is defined as the concentration of hydrogen ions [H+] in moles. It is given by the formula pH = log10 Find the amount of hydrogen in a liter of milk that has a pH of 6.7. A. 0.00000042 mole B. 0.00000034 mole C. 0.00000020 mole D. 0.0000017 mole Example 2

  29. SCIENCE The pH of a substance is defined as the concentration of hydrogen ions [H+] in moles. It is given by the formula pH = log10 Find the amount of hydrogen in a liter of milk that has a pH of 6.7. A. 0.00000042 mole B. 0.00000034 mole C. 0.00000020 mole D. 0.0000017 mole Example 2

  30. Concept

  31. Power Property of Logarithms Given that log5 6 ≈ 1.1133, approximate the value of log5 216. log5 216 = log5 63 Replace 216 with 63. = 3 log5 6 Power Property ≈ 3(1.1133) or 3.3399 Replace log5 6 with 1.1133. Answer: Example 3

  32. Power Property of Logarithms Given that log5 6 ≈ 1.1133, approximate the value of log5 216. log5 216 = log5 63 Replace 216 with 63. = 3 log5 6 Power Property ≈ 3(1.1133) or 3.3399 Replace log5 6 with 1.1133. Answer:3.3399 Example 3

  33. Given that log4 6 ≈ 1.2925, what is the approximate value of log4 1296? A. 0.3231 B. 2.7908 C. 5.1700 D. 6.4625 Example 3

  34. Given that log4 6 ≈ 1.2925, what is the approximate value of log4 1296? A. 0.3231 B. 2.7908 C. 5.1700 D. 6.4625 Example 3

  35. Original equation Power Property Quotient Property Property of Equality for Logarithmic Functions Multiply each side by 5. Solve Equations Using Properties of Logarithms Solve 4 log2x – log2 5 = log2 125. x = 5 Take the 4th root of each side. Example 4

  36. ? 4 log2 5 – log2 5 = log2 125 ? log2 54 – log2 5 = log2 125 ? ? log2 53 = log2 125 Solve Equations Using Properties of Logarithms Answer: Check Substitute each value into the original equation. 4 log2x – log2 5 = log2 125 log2 125 = log2 125 Example 4

  37. ? 4 log2 5 – log2 5 = log2 125 ? log2 54 – log2 5 = log2 125 ? ? log2 53 = log2 125 Solve Equations Using Properties of Logarithms Answer: 5 Check Substitute each value into the original equation. 4 log2x – log2 5 = log2 125 log2 125 = log2 125 Example 4

  38. Solve 2 log3 (x – 2) – log3 6 = log3 150. A.x = 4 B.x = 18 C.x = 32 D.x = 144 Example 4

  39. Solve 2 log3 (x – 2) – log3 6 = log3 150. A.x = 4 B.x = 18 C.x = 32 D.x = 144 Example 4

  40. End of the Lesson

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