8-4 Properties of Logarithms Using the properties of logarithms

Objectives Using the Properties of Logarithms

Vocabulary For any positive numbers, M,N, and b: logbMN = logbM + logbN Product Property logb = logbM - logbN Quotient Property logbMx = x logbM Power Property M N

Identifying the Properties of Logarithms State the property or properties used to rewrite each expression. a. log 6 = log 2 + log 3 Product Property: log 6 = log (2•3) = log 2 + log 3 b. logb = 2 logb x – logb y x2 y Quotient Property: logb = logb x2 – logb y x2 y Power Property: logb x2 – logb y = 2 logb x – logb y

Simplifying Logarithms Write each logarithmic expression as a single logarithm. a. log4 64 – log4 16 log4 64 – log4 16 = log4 Quotient Property 64 16 = log4 4 or 1 Simplify. b. 6 log5 x + log5 y 6 log5 x + log5 y = log5 x6 + log5 y Power Property = log5 (x6y) Product Property So log4 64 – log4 16 = log4 4, and 6 log5 x + log2 y = log5 (x6y).

Expanding Logarithms Expand each logarithm. a. log7 t u log7 = log7 t – log7 u Quotient Property t u b. log(4p3) log(4p3) = log 4 + log p3 Product Property = log 4 + 3 log p Power Property

Real-World Example Manufacturers of a vacuum cleaner want to reduce its sound intensity to 40% of the original intensity. By how many decibels would the loudness be reduced? Relate: The reduced intensity is 40% of the present intensity. Define: Let l1 = present intensity. Let l2 = reduced intensity. Let L1 = present loudness. Let L2 = reduced loudness. Write: l2 = 0.04 l1 L1 = 10 log L2 = 10 log l1 l0 l2

Continued (continued) L1 – L2 = 10 log l1 l0 l2 – 10 log Find the decrease in loudness L1 – L2. = 10 log l1 l0 0.40l1 – 10 log Substitute l2 = 0.40l1. – 10 log 0.40 • Product Property = 10 log l1 l0 – 10 ( log 0.40 + log ) = 10 log l1 l0 – 10 log 0.40 – 10 log Distributive Property = –10 log 0.40 Combine like terms. 4.0 Use a calculator. The decrease in loudness would be about 4 decibels.

Homework 8-4 Pg 457 # 1, 2, 3, 4, 11, 12, 19, 20