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

A. A

B. B

C. C

D. D0% 0%0%0%

A. x = 7

B. x = 6

C. x = 5

D. x = 4

Solve log4 (x2 – 30) = log4 x.

Over Lesson 8–4

A. A

B. B

C. C

D. D0% 0%0%0%

A. x = 2

B. x = 1

C. x = –5

D. x = –10

Solve log5 (x2 – 2x) = log5 (–5x + 10).

Over Lesson 8–4

A. A

B. B

C. C

D. D0% 0%0%0%

A. {x | 0 < x < 27}

B. {x | 0 < x < 18}

C. {x | 0 < x < 9}

D. {x | 0 < x < 6}

Solve log3 x < 30.

Over Lesson 8–4

A. A

B. B

C. C

D. D0% 0%0%0%

A. {x | x < 3}

B. {x | x > 3}

C. {x | x < 2}

D. {x | x > 2}

Solve log9 (4x + 6) > log9 (x + 12).

Over Lesson 8–4

A. A

B. B

C. C

D. D0% 0%0%0%

Solve log7 (x + 3) ≥ log7 (6x – 2).

A. {x | –1 < x < 2}

B.

C.

D. {x | 1 < x < 2}

Over Lesson 8–4

A. A

B. B

C. C

D. D0% 0%0%0%

Which of the following is not a solution to the inequality log8 (x – 2) ≤ log8 (5x – 6)?

A. –1

B. –

C.

D. 3

__1

2

__3

4

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 2Inverse Property of Exponents and Logarithms

≈ 3 + 0.4307 or 3.4307Replace log5 2 with 0.4307.

Answer: Thus, log5 250 is approximately 3.4307.

A. A

B. B

C. C

D. D0% 0%0%0%

A. –3.415

B. 3.415

C. 5.5850

D. 6.5850

Given log2 3 ≈ 1.5850, what is the approximate value of log2 96?

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

Quotient Property

Substitute 5.5 for pH.

log101 = 0

Quotient Property

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.

H+

H+

H+

Quotient Property

5.5 = log101 – log1010–5.5 Quotient Property?

5.5 = 0 – (–5.5) Simplify.?

5.5 = 5.5

pH = 5.5

?H+ = 10–5.5

Check

A. A

B. B

C. C

D. D0% 0%0%0%

A. 0.00000042 mole

B. 0.00000034 mole

C. 0.00000020 mole

D. 0.0000017 mole

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.

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 6Power Property

≈3(1.1133) or 3.3399Replace log5 6 with 1.1133.

Answer: 3.3399

A. A

B. B

C. C

D. D0% 0%0%0%

A. 0.3231

B. 2.7908

C. 5.1700

D. 6.4625

Given that log4 6 ≈ 1.2925, what is the approximate value of log4 1296?

Solve Equations Using Properties of Logarithms

Multiply each side by 5.

Solve 4 log2 x – log2 5 = log2 125.Original equation

Power Property

Quotient Property

Property of Equality for Logarithmic Functions

x = 5Take the 4th root of each side.

Solve Equations Using Properties of Logarithms

Answer: 5

4 log2 x – log2 5 = log2 125

Check Substitute each value into the original equation.

?4 log2 5 – log2 5 = log2 125

log2 54 – log2 5 = log2 125

?

log2 53 = log2 125

?

log2 125 = log2 125

?

A. A

B. B

C. C

D. D0% 0%0%0%

A. x = 4

B. x = 18

C. x = 32

D. x = 144

Solve 2 log3 (x – 2) log3 6 = log3 25.