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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 Lesson Menu

Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to approximate the value of log3 32. B. 2.3885 C. 3.1547 D. 4 5-Minute Check 1

Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to approximate the value of log3 32. B. 2.3885 C. 3.1547 D. 4 5-Minute Check 1

Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to approximate the value of log3 . __ 1 2 A. –0.6309 B. 0.1577 C. 0.3155 0.4732 5-Minute Check 2

Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to approximate the value of log3 . __ 1 2 A. –0.6309 B. 0.1577 C. 0.3155 0.4732 5-Minute Check 2

Solve log5 6 + 3 log5 x = log5 48. A. 1 B. 2 C. 3 D. 4 5-Minute Check 3

Solve log5 6 + 3 log5 x = log5 48. A. 1 B. 2 C. 3 D. 4 5-Minute Check 3

Solve log2 (n + 4) + log2 n = 5. A. 10 B. 8 C. 6 D. 4 5-Minute Check 4

Solve log2 (n + 4) + log2 n = 5. A. 10 B. 8 C. 6 D. 4 5-Minute Check 4

Solve log6 16 – 2 log6 4 = log6 (x + 1) + log6 . __ 1 4 A. 2 B. 3 C. 3.5 D. 4 5-Minute Check 5

Solve log6 16 – 2 log6 4 = log6 (x + 1) + log6 . __ 1 4 A. 2 B. 3 C. 3.5 D. 4 5-Minute Check 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 5-Minute Check 6

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 5-Minute Check 6

Mathematical Practices 4 Model with mathematics. Content Standards A.CED.1 Create equations and inequalities in one variable and use them to solve problems. Mathematical Practices 4 Model with mathematics. CCSS

Solve exponential equations and inequalities using common logarithms. 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. Then/Now

common logarithm Change of Base Formula Vocabulary

A. Use a calculator to evaluate log 6 to the nearest ten-thousandth. Find Common Logarithms A. Use a calculator to evaluate log 6 to the nearest ten-thousandth. Keystrokes: ENTER LOG 6 .7781512504 Answer: Example 1

A. Use a calculator to evaluate log 6 to the nearest ten-thousandth. Find Common Logarithms A. Use a calculator to evaluate log 6 to the nearest ten-thousandth. Keystrokes: ENTER LOG 6 .7781512504 Answer: about 0.7782 Example 1

Find Common Logarithms B. Use a calculator to evaluate log 0.35 to the nearest ten-thousandth. Keystrokes: ENTER LOG .35 –.4559319556 Answer: Example 1

Find Common Logarithms B. Use a calculator to evaluate log 0.35 to the nearest ten-thousandth. Keystrokes: ENTER LOG .35 –.4559319556 Answer: about –0.4559 Example 1

A. Which value is approximately equivalent to log 5? B. 0.6990 C. 5.0000 D. 100,000.0000 Example 1

A. Which value is approximately equivalent to log 5? B. 0.6990 C. 5.0000 D. 100,000.0000 Example 1

B. Which value is approximately equivalent to log 0.62? D. 4.1687 Example 1

B. Which value is approximately equivalent to log 0.62? D. 4.1687 Example 1

Solve Logarithmic Equations 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? Original equation Example 2

Replace L with 125 and m with 1. Solve Logarithmic Equations Replace L with 125 and m with 1. Divide each side by 10 and simplify. Exponential form Use a calculator. I ≈ 3.162 × 1012 Answer: Example 2

Replace L with 125 and m with 1. Solve Logarithmic Equations Replace L with 125 and m with 1. Divide each side by 10 and simplify. Exponential form Use a calculator. I ≈ 3.162 × 1012 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. Example 2

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? A. 1,585,000,000 times the minimum intensity B. 1,629,000,000 times the minimum intensity C. 1,912,000,000 times the minimum intensity D. 2,788,000,000 times the minimum intensity Example 2

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? A. 1,585,000,000 times the minimum intensity B. 1,629,000,000 times the minimum intensity C. 1,912,000,000 times the minimum intensity D. 2,788,000,000 times the minimum intensity Example 2

Solve 5x = 62. Round to the nearest ten-thousandth. Solve Exponential Equations Using Logarithms Solve 5x = 62. Round to the nearest ten-thousandth. 5x = 62 Original equation log 5x = log 62 Property of Equality for Logarithms x log 5 = log 62 Power Property of Logarithms Divide each side by log 5. x ≈ 2.5643 Use a calculator. Answer: Example 3

Solve 5x = 62. Round to the nearest ten-thousandth. Solve Exponential Equations Using Logarithms Solve 5x = 62. Round to the nearest ten-thousandth. 5x = 62 Original equation log 5x = log 62 Property of Equality for Logarithms x log 5 = log 62 Power Property of Logarithms Divide each side by log 5. x ≈ 2.5643 Use a calculator. Answer: about 2.5643 Example 3

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.  Example 3

What is the solution to the equation 3x = 17? A. x = 0.3878 B. x = 2.5713 C. x = 2.5789 D. x = 5.6667 Example 3

What is the solution to the equation 3x = 17? A. x = 0.3878 B. x = 2.5713 C. x = 2.5789 D. x = 5.6667 Example 3

Solve 37x > 25x – 3. Round to the nearest ten-thousandth. 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 Example 4

7x log 3 > 5x log 2 – 3 log 2 Distributive Property 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 Example 4

Divide each side by 7 log 3 – 5 log 2. Solve Exponential Inequalities Using Logarithms Divide each side by 7 log 3 – 5 log 2. Use a calculator. x > –0.4922 Simplify. Example 4

37x > 25x – 3 Original inequality Solve Exponential Inequalities Using Logarithms Check: Test x = 0. 37x > 25x – 3 Original inequality ? 37(0) > 25(0) – 3 Replace x with 0. ? 30 > 2–3 Simplify. Negative Exponent Property  Answer: Example 4

37x > 25x – 3 Original inequality Solve Exponential Inequalities Using Logarithms Check: Test x = 0. 37x > 25x – 3 Original inequality ? 37(0) > 25(0) – 3 Replace x with 0. ? 30 > 2–3 Simplify. Negative Exponent Property  Answer: The solution set is {x | x > –0.4922}. Example 4

What is the solution to 53x < 10x – 2? A. {x | x > –1.8233} B. {x | x < 0.9538} C. {x | x > –0.9538} D. {x | x < –1.8233} Example 4

What is the solution to 53x < 10x – 2? A. {x | x > –1.8233} B. {x | x < 0.9538} C. {x | x > –0.9538} D. {x | x < –1.8233} Example 4

Concept

Change of Base Formula Express log5 140 in terms of common logarithms. Then round to the nearest ten-thousandth. Change of Base Formula Use a calculator. Answer: Example 5

Answer: The value of log5 140 is approximately 3.0704. Change of Base Formula Express log5 140 in terms of common logarithms. Then round to the nearest ten-thousandth. Change of Base Formula Use a calculator. Answer: The value of log5 140 is approximately 3.0704. Example 5

What is log5 16 expressed in terms of common logarithms? B. C. D. Example 5

What is log5 16 expressed in terms of common logarithms? B. C. D. Example 5

End of the Lesson