1 / 40

Properties of Logarithms: Examples and Solutions

Learn how to use the product, quotient, and power properties of logarithms to simplify expressions and solve equations. This lesson includes real-world examples and practice problems.

josed
Download Presentation

Properties of Logarithms: Examples and Solutions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related