Five-Minute Check (over Lesson 8–4)
Then/Now
Key Concept: Product Property of Logarithms
Example 1: Use the Product Property
Key Concept: Quotient Property of Logarithms
Example 2: Real-World Example: Quotient Property
Key Concept: Power Property of Logarithms
Example 3: Power Property of Logarithms
Example 4: Solve Equations Using Properties of Logarithms
Over Lesson 8–4
Solve log4 (x2 – 30) = log4 x.
A. x = 7
B. x = 6
0%
A
B
C
0%
D
D
A
0%
B
D. x = 4
C
C. x = 5
A.
B.
C.
0%
D.
Over Lesson 8–4
Solve log5 (x2 – 2x) = log5 (–5x + 10).
A. x = 2
B. x = 1
0%
A
B
C
0%
D
D
A
0%
B
D. x = –10
C
C. x = –5
A.
B.
C.
0%
D.
Over Lesson 8–4
Solve log3 x < 30.
A. {x | 0 < x < 27}
B. {x | 0 < x < 18}
0%
A
B
C
0%
D
D
A
0%
B
D. {x | 0 < x < 6}
C
C. {x | 0 < x < 9}
A.
B.
C.
0%
D.
Over Lesson 8–4
Solve log9 (4x + 6) > log9 (x + 12).
A. {x | x < 3}
B. {x | x > 3}
0%
A
B
C
0%
D
D
A
0%
B
D. {x | x > 2}
C
C. {x | x < 2}
A.
B.
C.
0%
D.
Over Lesson 8–4
Solve log7 (x + 3) ≥ log7 (6x – 2).
A. {x | –1 < x < 2}
B.
0%
A
B
C
0%
D
D
A
0%
B
D. {x | 1 < x < 2}
C
C.
A.
B.
C.
0%
D.
Over Lesson 8–4
Which of the following is not a solution to the
inequality log8 (x – 2) ≤ log8 (5x – 6)?
A. –1
3
C. __
4
A
0%
0%
B
D. 3
A.
B.
C.
0%
D.
A
B
C
0%
D
D
2
C
1
B. – __
You evaluated logarithmic expressions and
solved logarithmic equations. (Lesson 8–4)
• Simplify and evaluate expressions using the
properties of logarithms.
• Solve logarithmic equations using the
properties of logarithms.
Use the Product Property
Use log5 2 ≈ 0.4307 to approximate the value of
log5 250.
log5 2 = 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.
Given log2 3 ≈ 1.5850, what is the approximate value
of log2 96?
A. –3.415
A
0%
0%
B
D. 6.5850
A
B
C
0%
D
D
C. 5.5850
A.
B.
C.
0%
D.
C
B. 3.415
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.
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
Quotient Property
H+
H+
H+
Simplify.
Multiply each side by –1.
Definition of logarithm
Answer: There are 10–5.5, or about 0.0000032, mole of
hydrogen in a liter of this rain.
Quotient Property
Check
pH = 5.5
?
?
5.5 = log101 – log1010–5.5
?
5.5 = 0 – (–5.5)
5.5 = 5.5 
H+ = 10–5.5
Quotient Property
Simplify.
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
D. 0.0000017 mole
0%
0%
B
C. 0.00000020 mole
A
B
C
0%
D
D
B. 0.00000034 mole
A.
B.
C.
0%
D.
C
A. 0.00000042 mole
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
Given that log4 6 ≈ 1.2925, what is the approximate
value of log4 1296?
A. 0.3231
B. 2.7908
0%
B
A
0%
A
B
C
0%
D
D
D. 6.4625
C
C. 5.1700
A.
B.
C.
0%
D.
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.
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
?
3 ?
log2 5 = log2 125
log2 125 = log2 125
Solve 2 log3 (x – 2) log3 6 = log3 25.
A. x = 4
A
0%
0%
B
D. x = 144
A
B
C
0%
D
D
C. x = 32
A.
B.
C.
0%
D.
C
B. x = 18