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Presentation transcript:

Concept

Use log5 2 ≈ 0.4307 to approximate the value of log5 250. 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

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

Concept

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

Understand The formula for finding pH and the pH of the rain is given. 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 Original equation Substitute 5.5 for pH. Quotient Property log101 = 0 Example 2

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

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

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

Concept

Given that log5 6 ≈ 1.1133, approximate the value of log5 216. 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

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

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

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

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