Use a calculator to evaluate each expression to the nearest ten-thousandth. 17. log 11. SOLUTION: KEYSTROKES: LOG 11 ENTER 1.041392685. 19. log 8.2. S...

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19. log 8.2 SOLUTION: KEYSTROKES: LOG 8 . 2 ENTER 0.913813852

21. log 0.04 SOLUTION: KEYSTROKES: LOG 0 . 0 4 ENTER –1.39794001

Solve each equation. Round to the nearest ten-thousandth. x

23. 8 = 40 SOLUTION:

The solution is about 1.7740. a − 4

25. 2.9

= 8.1

SOLUTION:

The solution is about 5.9647.

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27.

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7-6 Common Logarithms The solution is about 1.7740. a − 4

25. 2.9

= 8.1

SOLUTION:

The solution is about 5.9647. 27. SOLUTION:

The solution is about 1.1691. Solve each inequality. Round to the nearest ten-thousandth. 3n

29. 6

> 36

SOLUTION:

The solution region is {n | n > 0.6667}. y − 1

31. 3

y

≤4

SOLUTION: eSolutions Manual - Powered by Cognero

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7-6 Common Logarithms The solution region is {n | n > 0.6667}. y − 1

31. 3

y

≤4

SOLUTION:

The solution region is

.

Express each logarithm in terms of common logarithms. Then approximate its value to the nearest tenthousandth. 33. log7 18 SOLUTION:

35. log2 16 SOLUTION:

37. log3 11 SOLUTION:

39. PETS The number n of pet owners in thousands after t years can be modeled by n = 35[log4 (t + 2)]. Let t = 0 represent 2000. Use the Change of Base Formula to solve the following questions. a. How many pet owners were there in 2010? b. How long until there are 80,000 pet owners? When will this occur? eSolutions Manual - Powered by Cognero SOLUTION:

a. The value of t at 2010 is 10. Substitute 10 for t in the equation and evaluate.

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SOLUTION: 7-6 Common Logarithms 39. PETS The number n of pet owners in thousands after t years can be modeled by n = 35[log4 (t + 2)]. Let t = 0 represent 2000. Use the Change of Base Formula to solve the following questions. a. How many pet owners were there in 2010? b. How long until there are 80,000 pet owners? When will this occur? SOLUTION: a. The value of t at 2010 is 10. Substitute 10 for t in the equation and evaluate.

There will be 62,737 pet owners in 2010.

b. Substitute 80 for n and solve for t.

In 2022, there will be 80,000 pet owners. Solve each equation or inequality. Round to the nearest ten-thousandth. x

41. 3 = 40 SOLUTION:

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The solution is about 3.3578.

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7-6 Common Logarithms In 2022, there will be 80,000 pet owners. Solve each equation or inequality. Round to the nearest ten-thousandth. x

41. 3 = 40 SOLUTION:

The solution is about 3.3578. n+2

43. 4

= 14.5

SOLUTION:

The solution is about –0.0710. n − 3

45. 7.4

= 32.5

SOLUTION:

The solution is about 4.7393. x

47. 5 ≥ 42 SOLUTION:

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7-6 Common Logarithms The solution is about 4.7393. x

47. 5 ≥ 42 SOLUTION:

The solution region is {x | x ≥ 2.3223}. 4x

49. 3 ≤ 72 SOLUTION:

The solution region is {x | x ≤ 0.9732}. p

5 − p

51. 6 ≤ 13

SOLUTION:

The solution region is {p | p ≤ 2.9437}. Express each logarithm in terms of common logarithms. Then approximate its value to the nearest tenthousandth. 53. log4 12 SOLUTION: eSolutions Manual - Powered by Cognero

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7-6 Common Logarithms The solution region is {p | p ≤ 2.9437}. Express each logarithm in terms of common logarithms. Then approximate its value to the nearest tenthousandth. 53. log4 12 SOLUTION:

55. log8 2 SOLUTION:

2

57. log5 (2.7)

SOLUTION:

59. MUSIC A musical cent is a unit in a logarithmic scale of relative pitch or intervals. One octave is equal to 1200 cents. The formula

can be used to determine the difference in cents between two notes with

frequencies a and b. a. Find the interval in cents when the frequency changes from 443 Hertz (Hz) to 415 Hz. b. If the interval is 55 cents and the beginning frequency is 225 Hz, find the final frequency. SOLUTION: a. Substitute 443 and 415for a and b then evaluate.

The interval is 113.03 cents.

b. Substitute 55 and 225 for n and a then solve for b.

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SOLUTION: 7-6 Common Logarithms 59. MUSIC A musical cent is a unit in a logarithmic scale of relative pitch or intervals. One octave is equal to 1200 can be used to determine the difference in cents between two notes with

cents. The formula

frequencies a and b. a. Find the interval in cents when the frequency changes from 443 Hertz (Hz) to 415 Hz. b. If the interval is 55 cents and the beginning frequency is 225 Hz, find the final frequency. SOLUTION: a. Substitute 443 and 415for a and b then evaluate.

The interval is 113.03 cents.

b. Substitute 55 and 225 for n and a then solve for b.

The final frequency is 218. Solve each equation. Round to the nearest ten-thousandth. 61. SOLUTION:

The solutions are eSolutions Manual - Powered by Cognero

63.

. Page 8

7-6 Common Logarithms The final frequency is 218. Solve each equation. Round to the nearest ten-thousandth. 61. SOLUTION:

The solutions are

.

63. SOLUTION:

The solution is 3.5. 65. SOLUTION:

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7-6 Common Logarithms The solution is 3.5. 65. SOLUTION:

The solution is about –3.8188.

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