Over Lesson 7 5 Today is day 55

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Over Lesson 7– 5 • Today is day 55, section 7. 6 Common Logarithms

Over Lesson 7– 5 • Today is day 55, section 7. 6 Common Logarithms • Do NOW (one of my favorites!!) • Correct homework and Frames • Homework tonight: Day 55, Frames

Over Lesson 7– 5 Solve log 5 6 + 3 log 5 x =

Over Lesson 7– 5 Solve log 5 6 + 3 log 5 x = log 5 48. A. 1 B. 2 C. 3 D. 4

Over Lesson 7– 5 Solve log 5 6 + 3 log 5 x =

Over Lesson 7– 5 Solve log 5 6 + 3 log 5 x = log 5 48. A. 1 B. 2 C. 3 D. 4

Over Lesson 7– 5 Solve log 2 (n + 4) + log 2 n

Over Lesson 7– 5 Solve log 2 (n + 4) + log 2 n = 5. A. 10 B. 8 C. 6 D. 4

Over Lesson 7– 5 Solve log 2 (n + 4) + log 2 n

Over Lesson 7– 5 Solve log 2 (n + 4) + log 2 n = 5. A. 10 B. 8 C. 6 D. 4

Over Lesson 7– 5 1 Solve log 6 16 – 2 log 6 4

Over Lesson 7– 5 1 Solve log 6 16 – 2 log 6 4 = log 6 (x + 1) + log 6 __. 4 A. 2 B. 3 C. 3. 5 D. 4

Over Lesson 7– 5 1 Solve log 6 16 – 2 log 6 4

Over Lesson 7– 5 1 Solve log 6 16 – 2 log 6 4 = log 6 (x + 1) + log 6 __. 4 A. 2 B. 3 C. 3. 5 D. 4

Over Lesson 7– 5 Which of the following equations is false? A. log 8

Over Lesson 7– 5 Which of the following equations is false? A. log 8 m 5 = 5 log 8 m B. loga 6 – loga 3 = loga 2 C. D. logb 2 x = logb 2 + logb x

Over Lesson 7– 5 Which of the following equations is false? A. log 8

Over Lesson 7– 5 Which of the following equations is false? A. log 8 m 5 = 5 log 8 m B. loga 6 – loga 3 = loga 2 C. D. logb 2 x = logb 2 + logb x

You simplified expressions and solved equations using properties of logarithms. • Solve exponential equations

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

• common logarithm • Change of Base Formula

Find Common Logarithms A. Use a calculator to evaluate log 6 to the nearest

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

Find Common Logarithms A. Use a calculator to evaluate log 6 to the nearest

Find Common Logarithms A. Use a calculator to evaluate log 6 to the nearest ten-thousandth. Keystrokes: LOG Answer: about 0. 7782 6 ENTER . 7781512504

Find Common Logarithms B. Use a calculator to evaluate log 0. 35 to the

Find Common Logarithms B. Use a calculator to evaluate log 0. 35 to the nearest ten-thousandth. Keystrokes: LOG Answer: . 35 ENTER –. 4559319556

Find Common Logarithms B. Use a calculator to evaluate log 0. 35 to the

Find Common Logarithms B. Use a calculator to evaluate log 0. 35 to the nearest ten-thousandth. Keystrokes: LOG . 35 ENTER Answer: about – 0. 4559 –. 4559319556

Solve Logarithmic Equations JET ENGINES The loudness L, in decibels, of a sound is

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

Solve Logarithmic Equations Replace L with 125 and m with 1. Divide each side

Solve Logarithmic Equations Replace L with 125 and m with 1. Divide each side by 10 and simplify. Exponential form I ≈ 3. 162 × 1012 Answer: Use a calculator.

Solve Logarithmic Equations Replace L with 125 and m with 1. Divide each side

Solve Logarithmic Equations Replace L with 125 and m with 1. Divide each side by 10 and simplify. Exponential form I ≈ 3. 162 × 1012 Use a calculator. 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.

DEMOLITION The loudness L, in decibels, of a sound is where I is the

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

DEMOLITION The loudness L, in decibels, of a sound is where I is the

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

Solve Exponential Equations Using Logarithms Solve 5 x = 62. Round to the nearest

Solve Exponential Equations Using Logarithms Solve 5 x = 62. Round to the nearest ten-thousandth. 5 x = 62 Original equation log 5 x = 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 Answer: Use a calculator.

Solve Exponential Equations Using Logarithms Solve 5 x = 62. Round to the nearest

Solve Exponential Equations Using Logarithms Solve 5 x = 62. Round to the nearest ten-thousandth. 5 x = 62 Original equation log 5 x = 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 Answer: about 2. 5643 Use a calculator.

Solve Exponential Equations Using Logarithms Check You can check this answer by using a

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.

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

What is the solution to the equation 3 x = 17? A. x = 0. 3878 B. x = 2. 5713 C. x = 2. 5789 D. x = 5. 6667

Change of Base Formula Express log 5 140 in terms of common logarithms. Then

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

Change of Base Formula Express log 5 140 in terms of common logarithms. Then

Change of Base Formula Express log 5 140 in terms of common logarithms. Then round to the nearest ten-thousandth. Change of Base Formula Use a calculator. Answer: The value of log 5 140 is approximately 3. 0704.

• Solve: • 35 y = 7 y-2 • Answer: y = -1.

• Solve: • 35 y = 7 y-2 • Answer: y = -1. 0972