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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

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.

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).

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.

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).

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}

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 __

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.

Concept

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

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

Concept

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.

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

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

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

A.A B.B C.C D.D Example 2 A mole B mole C mole D 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.

Concept

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

A.A B.B C.C D.D Example 3 A B C D Given that log 4 6 ≈ , what is the approximate value of log ?

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

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

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.

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