1. Splash Screen

2. Lesson Menu Five-Minute Check (over Lesson 8–4) 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

3. Over Lesson 8–4 A.A B.B C.C D.D 5-Minute Check 1 A.x = 7 B.x = 6 C.x = 5 D.x = 4 Solve log 4 (x 2 – 30) = log 4 x.

4. Over Lesson 8–4 A.A B.B C.C D.D 5-Minute Check 2 A.x = 2 B.x = 1 C.x = –5 D.x = –10 Solve log 5 (x 2 – 2x) = log 5 (–5x + 10).

5. Over Lesson 8–4 A.A B.B C.C D.D 5-Minute Check 3 A.{x | 0 < x < 27} B.{x | 0 < x < 18} C.{x | 0 < x < 9} D.{x | 0 < x < 6} Solve log 3 x < 30.

6. Over Lesson 8–4 A.A B.B C.C D.D 5-Minute Check 4 A.{x | x < 3} B.{x | x > 3} C.{x | x < 2} D.{x | x > 2} Solve log 9 (4x + 6) > log 9 (x + 12).

7. Over Lesson 8–4 A.A B.B C.C D.D 5-Minute Check 5 Solve log 7 (x + 3) ≥ log 7 (6x – 2). A.{x | –1 < x < 2} B. C. D.{x | 1 < x < 2}

8. Over Lesson 8–4 A.A B.B C.C D.D 5-Minute Check 6 Which of the following is not a solution to the inequality log 8 (x – 2) ≤ log 8 (5x – 6)? A.–1 B. – C. D.3 __ 1 2 3 4

9. Then/Now You evaluated logarithmic expressions and solved logarithmic equations. (Lesson 8–4) Simplify and evaluate expressions using the properties of logarithms. Solve logarithmic equations using the properties of logarithms.

10. Concept

11. Example 1 Use the Product Property Use log 5 2 ≈ 0.4307 to approximate the value of log 5 250. log 5 2 = log 5 (5 3 ● 2)Replace 250 with 5 3 ● 2. = log 5 5 3 + log 5 2Product Property = 3 + log 5 2Inverse Property of Exponents and Logarithms ≈ 3 + 0.4307 or 3.4307Replace log 5 2 with 0.4307. Answer: Thus, log 5 250 is approximately 3.4307.

12. A.A B.B C.C D.D Example 1 A.–3.415 B.3.415 C.5.5850 D.6.5850 Given log 2 3 ≈ 1.5850, what is the approximate value of log 2 96?

13. Concept

14. Example 2 Quotient Property SCIENCE The 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.

15. Example 2 Quotient Property UnderstandThe formula for finding pH and the pH of the rain is given. PlanWrite the equation. Then, solve for [H + ]. Solve Original equation Quotient Property Substitute 5.5 for pH. log 10 1 = 0

16. Example 2 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. H+H+ H+H+ H+H+

17. Example 2 Quotient Property 5.5= log 10 1 – log 10 10 –5.5 Quotient Property ? 5.5= 0 – (–5.5)Simplify. ? 5.5= 5.5  pH = 5.5 ? H + = 10 –5.5 Check

18. A.A B.B C.C D.D Example 2 A.0.00000042 mole B.0.00000034 mole C.0.00000020 mole D.0.0000017 mole 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.

19. Concept

20. Example 3 Power Property of Logarithms Given that log 5 6 ≈ 1.1133, approximate the value of log 5 216. log 5 216=log 5 6 3 Replace 216 with 6 3. =3 log 5 6Power Property ≈3(1.1133) or 3.3399Replace log 5 6 with 1.1133. Answer: 3.3399

21. A.A B.B C.C D.D Example 3 A.0.3231 B.2.7908 C.5.1700 D.6.4625 Given that log 4 6 ≈ 1.2925, what is the approximate value of log 4 1296?

22. Example 4 Solve Equations Using Properties of Logarithms Multiply each side by 5. Solve 4 log 2 x – log 2 5 = log 2 125. Original equation Power Property Quotient Property Property of Equality for Logarithmic Functions x=5Take the 4th root of each side.

23. Example 4 Solve Equations Using Properties of Logarithms Answer: 5 4 log 2 x – log 2 5 = log 2 125 Check Substitute each value into the original equation. ? 4 log 2 5 – log 2 5 = log 2 125 log 2 5 4 – log 2 5 = log 2 125 ? log 2 5 3 = log 2 125 ? log 2 125 = log 2 125  ?

24. A.A B.B C.C D.D Example 4 A.x = 4 B.x = 18 C.x = 32 D.x = 144 Solve 2 log 3 (x – 2) log 3 6 = log 3 25.

25. End of the Lesson