Five-Minute Check (over Lesson 7–5)

CCSS

Then/Now

New Vocabulary

Example 1: Find Common Logarithms

Example 2: Real-World Example: Solve Logarithmic Equations

Example 3: Solve Exponential Equations Using Logarithms

Example 4: Solve Exponential Inequalities Using Logarithms

Key Concept: Change of Base Formula

Example 5: Change of Base Formula

Over Lesson 7–5

A. 1.9864

B. 2.3885

C. 3.1547

D. 4

Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to approximate the value of log3 32.

Over Lesson 7–5

A. –0.6309

B. 0.1577

C. 0.3155

D. 0.4732

Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to approximate the value of log3 .

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Over Lesson 7–5

A. 1

B. 2

C. 3

D. 4

Solve log5 6 + 3 log5 x = log5 48.

Over Lesson 7–5

A. 10

B. 8

C. 6

D. 4

Solve log2 (n + 4) + log2 n = 5.

Over Lesson 7–5

A. 2

B. 3

C. 3.5

D. 4

Solve log6 16 – 2 log6 4 = log6 (x + 1) + log6 .__14

Over Lesson 7–5

Which of the following equations is false?

A. log8 m5 = 5 log8 m

B. loga 6 – loga 3 = loga 2

C.

D. logb 2x = logb 2 + logb x

Content Standards

A.CED.1 Create equations and inequalities in one variable and use them to solve problems.

Mathematical Practices

4 Model with mathematics.

You simplified expressions and solved equations using properties of logarithms.

• Solve exponential equations and inequalities using common logarithms.

• Evaluate logarithmic expressions using the Change of Base Formula.

• common logarithm

• Change of Base Formula

Find Common Logarithms

A. Use a calculator to evaluate log 6 to the nearest ten-thousandth.

Answer: about 0.7782

Keystrokes: ENTERLOG 6 .7781512504

Find Common Logarithms

B. Use a calculator to evaluate log 0.35 to the nearest ten-thousandth.

Answer: about –0.4559

Keystrokes: ENTERLOG .35 –.4559319556

A. 0.3010

B. 0.6990

C. 5.0000

D. 100,000.0000

A. Which value is approximately equivalent to log 5?

A. –0.2076

B. 0.6200

C. 1.2076

D. 4.1687

B. Which value is approximately equivalent to log 0.62?

Solve Logarithmic Equations

Original equation

JET ENGINES The loudness L, in decibels, of a

sound is where I is the intensity of

the sound and m is the minimum intensity of sound detectable by the human ear. The sound of a jet engine can reach a loudness of 125 decibels. How many times the minimum intensity of audible sound is this, if m is defined to be 1?

Solve Logarithmic Equations

Exponential form

Answer: The sound of a jet engine is approximately 3 × 1012 or 3 trillion times the minimum intensity of sound detectable by the human ear.

Use a calculator.I ≈ 3.162 × 1012

Replace L with 125 and m with 1.

Divide each side by 10 and simplify.

A. 1,585,000,000 times the minimum intensityB. 1,629,000,000 times the minimum intensityC. 1,912,000,000 times the minimum intensityD. 2,788,000,000 times the minimum intensity

DEMOLITION The loudness L, in decibels, of a

sound is where I is the intensity of the

sound and m is the minimum intensity of sound detectable by the human ear. Refer to Example 2. The sound of the demolition of an old building can reach a loudness of 92 decibels. How many times the minimum intensity of audible sound is this, if m is defined to be 1?

Solve Exponential Equations Using Logarithms

Solve 5x = 62. Round to the nearest ten-thousandth.

5x = 62 Original equation

log 5x = log 62Property of Equality for Logarithms

x log 5= log 62Power Property of Logarithms

Answer: about 2.5643

x ≈ 2.5643 Use a calculator.

Divide each side by log 5.

Solve Exponential Equations Using Logarithms

Check You can check this answer by using a calculator or by using estimation. Since 52 = 25 and 53 = 125, the value of x is between 2 and 3. Thus, 2.5643 is a reasonable solution.

A. x = 0.3878

B. x = 2.5713

C. x = 2.5789

D. x = 5.6667

What is the solution to the equation 3x = 17?

Solve Exponential Inequalities Using Logarithms

Solve 37x > 25x – 3. Round to the nearest ten-thousandth.

37x

> 25x – 3

Original inequality

log 37x

> log 25x – 3

Property of Inequality for Logarithmic Functions

7x log 3

> (5x – 3) log 2

Power Property of Logarithms

Solve Exponential Inequalities Using Logarithms

7x log 3

> 5x log 2 – 3 log 2

Distributive Property

7x log 3 – 5x log 2

> – 3 log 2

Subtract 5x log 2 from each side.

x(7 log 3 – 5 log 2) > –3 log 2

Distributive Property

x > –0.4922 Simplify.

Solve Exponential Inequalities Using Logarithms

Use a calculator.

Divide each side by 7 log 3 – 5 log 2.

Solve Exponential Inequalities Using Logarithms

Check: Test x = 0.

37x > 25x – 3 Original inequality

Answer: The solution set is {x | x > –0.4922}.

?37(0)> 25(0) – 3 Replace x with 0.?30 > 2–3 Simplify.

Negative Exponent Property

A. {x | x > –1.8233}

B. {x | x < 0.9538}

C. {x | x > –0.9538}

D. {x | x < –1.8233}

What is the solution to 53x < 10x – 2?

Change of Base Formula

Express log5 140 in terms of common logarithms. Then round to the nearest ten-thousandth.

Answer: The value of log5 140 is approximately 3.0704.

Use a calculator.

Change of Base Formula

What is log5 16 expressed in terms of common logarithms?

A.

B.

C.

D.