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Published byMadeline Clay Modified over 4 years ago

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**Five-Minute Check (over Lesson 9-3) Main Ideas and 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 4 Menu

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**Solve exponential equations and inequalities using common logarithms.**

Evaluate logarithmic expressions using the Change of Base Formula. common logarithm Change of Base Formula Lesson 4 MI/Vocab

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**Find Common Logarithms**

A. Use a calculator to evaluate log 6 to four decimal places. Keystrokes: ENTER LOG 6 Answer: about Lesson 4 Ex1

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**Find Common Logarithms**

B. Use a calculator to evaluate log 0.35 to four decimal places. Keystrokes: ENTER LOG .35 – Answer: about –0.4559 Lesson 4 Ex1

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**A. Which value is approximately equivalent to log 5?**

B C D. 100, A B C D Lesson 4 CYP1

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**B. Which value is approximately equivalent to log 0.62?**

D A B C D Lesson 4 CYP1

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**Solve Logarithmic Equations**

EARTHQUAKE The amount of energy E, in ergs, that an earthquake releases is related to its Richter scale magnitude M by the equation log E = M. The San Fernando Valley earthquake of 1994 measured 6.6 on the Richter scale. How much energy did this earthquake release? log E = M Write the formula. log E = (6.6) Replace M with 6.6. log E = 21.7 Simplify. 10log E = Write each side using 10 as a base. Lesson 4 Ex2

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**Solve Logarithmic Equations**

E = Inverse Property of Exponents and Logarithms E ≈ 5.01 × 1021 Use a calculator. Answer: The amount of energy released was about 5.01 × 1021 ergs. Lesson 4 Ex2

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EARTHQUAKE The amount of energy E, in ergs, that an earthquake releases is related to its Richter scale magnitude M by the equation log E = M. In 1999 an earthquake in Turkey measured 7.4 on the Richter scale. How much energy did this earthquake release? A. –7.29 ergs B. –2.93 ergs C ergs D × 1022 ergs A B C D Lesson 4 CYP2

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**Solve Exponential Equations Using Logarithms**

Solve 5x = 62. 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 ≈ Use a calculator. Answer: Lesson 4 Ex3

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**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, is a reasonable solution. Lesson 4 Ex3

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

B C D A B C D Lesson 4 CYP3

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**Solve Exponential Inequalities Using Logarithms**

Solve 27x > 35x – 3. 27x > 35x – 3 Original inequality log 27x > log 35x – 3 Property of Inequality for Logarithmic Functions 7x log 2 > (5x – 3) log 3 Power Property of Logarithms 7x log 2 > 5x log 3 – 3 log 3 Distributive Property 7x log 2 – 5x log 3 > – 3 log 3 Subtract 5x log 3 from each side. Lesson 4 Ex4

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**Solve Exponential Inequalities Using Logarithms**

x(7 log 2 – 5 log 3) > –3 log 3 Distributive Property Divide each side by 7 log 2 – 5 log 3. Switch > to < because 7 log 2 – 5 log 3 is negative. Use a calculator. Simplify. Lesson 4 Ex4

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**Solve Exponential Inequalities Using Logarithms**

Check: Test x = 0. 27x > 35x – 3 Original inequality ? 27(0) > 35(0) – 3 Replace x with 0. ? 20 > 3–3 Simplify. Negative Exponent Property Answer: The solution set is {x | x < }. Lesson 4 Ex4

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

A. {x | x > –1.8233} B. {x | x < } C. {x | x > –0.9538} D. {x | x < –1.8233} A B C D Lesson 4 CYP4

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Lesson 4 KC1

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**Answer: The value of log3 18 is approximately 2.6309.**

Change of Base Formula Express log3 18 in terms of common logarithms. Then approximate its value to four decimal places. Change of Base Formula Use a calculator. Answer: The value of log3 18 is approximately Lesson 4 Ex5

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**What is log5 16 expressed in terms of common logarithms and approximated to four decimal places?**

B. C. D. A B C D Lesson 4 CYP5

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End of Lesson 4

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