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
M
logb N
= logbM - logbN
logbMx = x logbM
Product Property
Quotient Property
Power Property
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
2
b. logb x = 2 logb x – logb y
y
2
Quotient Property: logb x = logb x2 – logb y
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 64
16
= log4 4 or 1
Quotient Property
Simplify.
b. 6 log5 x + log5 y
6 log5 x + log5 y = log5 x6 + log5 y
= log5 (x6y)
Power Property
Product Property
So log4 64 – log4 16 = log4 4, and 6 log5 x + log2 y = log5 (x6y).
Expanding Logarithms
Expand each logarithm.
a. log7
log7
t
u
t = log t – log u
7
7
u
Quotient Property
b. log(4p3)
log(4p3) = log 4 + log p3
= log 4 + 3 log p
Product Property
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 l1
l0
L2 = 10 log l2
l0
Continued
(continued)
L1 – L2 = 10 log l1 – 10 log l2
Find the decrease in loudness L1 – L2.
l0
l0
= 10 log l1 – 10 log0.40l1
Substitute l2 = 0.40l1.
l0
l0
= 10 log l1 – 10 log 0.40 • l1
l0
l0
= 10 log l1 – 10 ( log 0.40 + log l1 )
Product Property
l0
l0
= 10 log l1 – 10 log 0.40 – 10 log l1
Distributive Property
l0
l0
= –10 log 0.40
4.0
Combine like terms.
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