1. 8-4 Properties of Logarithms
Using the properties of logarithms

2. Objectives
Using the Properties of Logarithms

3. 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

4. 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

5. 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).

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

7. 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

8. 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 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.

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