1. Splash Screen

2. 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

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

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

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

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

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

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

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

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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. common logarithm
Change of Base Formula
Vocabulary

18. 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
Answer:
Example 1

19. 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
Answer: about
Example 1

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

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

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

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

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

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

26. 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

27. 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 ≈ × 1012
Answer:
Example 2

28. 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 ≈ × 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

29. 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

30. 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

31. 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 ≈ Use a calculator.
Answer:
Example 3

32. 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 ≈ Use a calculator.
Answer: about
Example 3

33. 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. 
Example 3

34. What is the solution to the equation 3x = 17?
A. x =
B. x =
C. x =
D. x =
Example 3

35. What is the solution to the equation 3x = 17?
A. x =
B. x =
C. x =
D. x =
Example 3

36. 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

37. 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

38. 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 > – Simplify.
Example 4

39. 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

40. 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

41. 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}
Example 4

42. 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}
Example 4

43. Concept

44. 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

45. 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
Example 5

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

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

48. End of the Lesson