Properties of Logarithms
ALGEBRA 2 LESSON 8-4
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
x2
= 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
8-4
Properties of Logarithms
ALGEBRA 2 LESSON 8-4
Write each logarithmic expression as a single logarithm.
a. log4 64 – log4 16
log4 64 – log4 16 = log4 64
Quotient Property
16
= log4 4 or 1
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).
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Properties of Logarithms
ALGEBRA 2 LESSON 8-4
Expand each logarithm.
a. log7
t
u
log7
t
u
= log7 t – log7 u
Quotient Property
b. log(4p3)
log(4p3) = log 4 + log p3
Product Property
= log 4 + 3 log p
Power Property
8-4
Properties of Logarithms
ALGEBRA 2 LESSON 8-4
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
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Properties of Logarithms
ALGEBRA 2 LESSON 8-4
(continued)
L1 – L2 = 10 log l1 – 10 log l2
l0
Find the decrease in loudness L1 – L2.
l0
= 10 log l1 – 10 log 0.40l1
l0
Substitute l2 = 0.40l1.
l0
= 10 log l1 – 10 log 0.40 • l1
l0
= 10 log l1 – 10
l0
l0
( log 0.40 + log
l1
l0
)
= 10 log l1 – 10 log 0.40 – 10 log l1
l0
l0
= –10 log 0.40
4.0
Product Property
Distributive Property
Combine like terms.
Use a calculator.
The decrease in loudness would be about 4 decibels.
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