Solutions and Hints

by Brent M. Dingle

for the book:

Precalculus, Mathematics for Calculus 4^th Edition

by James Stewart, Lothar Redlin and Saleem Watson.

Problems 1.2:

1. 

2. 

3. 

4. 

5. 

6. 

For 7 to 14 most odds are left undone – see your book for answers.

Remember to change all radicals expressions to exponents and

look for common bases – for example if you see a 4 remember that 2² = 4.

Also remember 5³ = 125, 2⁶ = 4³ = 8² = 64, 2¹⁰ = 1024, 2³ = 8, 3³ = 27.

And there are probably others.

Some rules to remember are:

- anything to the zero power is 1.

- (x^A)^B = x^A*B

- (x^A)^1/A = x¹ = x

7. See your book

8. a)
   
   
   
   
   b)
   
   
   c)

9. See your text book.

10. a)
    
    b)
    
    c) Here we might think: Gee is -32 = anything to the fifth?
    And because the book is “nice” there is a solution:
    

11. a) recall 2² = 4 and 3² = 9
    b) recall 4⁴ = 256
    c) recall 2⁶ = 64

12. a) pretty tricky yes?
    
    
    b) recall 48 = 16*3
    
    
    
    c) recall 24 = 3 * 8 and 54 = 27 * 2
    
    So far so good, right? Now 81 is what to the fourth?
    And 16 is what to the fourth?
    
    
    
    

13. a)
    
    b) Recall –2⁵ = -32 and (-2)² = 4
    
    c) Recall 5³ = 125

14. a)
    
    b) Recall –27 = (-3)³ and 8 = 2³.
    
    
    c) Recall 25 = 5² and 64 = 8²

Problems 15 to 18 you just “plug in” the values for x, y and z.

x = 3, y = 4, and z = -1

15. (3² + 4²)^1/2 = (25)^1/2 = 5

16. (3³ + 14*4 + 2*(-1) )^1/4 = (27 + 56 – 2)^1/4 = 81^1/4 = (3⁴)^1/4 = 3

17. (9*3)^2/3 + (2*4)^2/3 + (-1)^2/3 = (3³)^2/3 + (2³)^2/3 + (-1²)^1/3 = 3²+2²+1 = 9+4+1 = 14

18. (3*4)^2*(-1) = (12)^-2 = 1 / (12²) = 1 / 144 » 0.00694444…

19. Notice 108 = 27*4 and 32 = 8*4
    So you get:

20. Notice 8 = 4*2 and 50 = 25*2 (you look for perfect squares to ‘pull out’)
    So you get:
    

21. Notice 245 = 49 * 5 and 125 = 25 * 5

22. Notice 54 = 27 * 2 and 16 = 8 * 2
    So you get:
    

For 23 to 40 remember these rules:

1. x^A*x^B = x^A+B
2. (x^A)^B = x^A*B
3. x^A / x^B = x^A-B
4. (x^Ay^B)^C = x^ACy^BC

23. Notice 9 + (-5) = 4

24. 3*4 = 12 and 2 + 5 = 7 è 12y⁷

25. 12*1/2 = 6, 2+5 = 7, and 4 + 1 = 5 è 6x⁷y⁵ .

26. Remember to distribute the power è (6y)³ = 6³y³ = 1296y³ .

27. 2⁴ = 16, 9 + 4 – 3 = 10 è 16x¹⁰ .

28. –3 – (-5) = 2, 4 – 5 = -1 è a²b^-1 = a² / b

29. see book è 4/b².

30. 2*(1/4)*16 = 8, 3+6 = 9, -1+4 = 3 è 8s⁹t³ .

31. see book, remember to distribute the powers è 64r⁷s

32. (2u²v³)³ = 8u⁶v⁹ and (3u³v)^-2 = (1/9)u^-6v^-2 .
    so (8u⁶v⁹)* (1/9)u^-6v^-2 = (8/9)u⁰v⁷ = (8/9)*1*v⁷ = (8/9)v⁷ .

33. see book, 6⁴ / 2 = 1296 / 2 = 648, (3*4)-5 = 7 è 648y⁷

34. you can do

35. see book è x³ / y

36. you can do

37. see book è (y²z⁹) / x⁵

38. 
    
    

39. see your book è s³ / (q⁷r⁶)

40. 
    
    
    
    

41 to 56 are similar to the 23 to 40 except they have fractional exponents.

The values of the variables must be assumed to be positive numbers as you cannot take the even root of negative numbers (e.g. (-3)^1/2 = imaginary number).

41. 2/3 + 1/5 = 13/15, see book è x^13/15 .

42. –2*5 = -10, ¾ + 3/2 = 9/4 è -10a^9/4 .

43. Be sure to distribute the exponents to the coefficients, see book è 16*b^9/10 .

44. 

45. Distribute the exponent, see book è 1 / (c^2/3d)

46. Distribute exponent, recall 4 = 2² è 8x⁹y¹² .

47. ¾ * 2/3 = 6/12, see book è y^1/2 .

48. 2/5 * -3/4 = -6/20 = -3/10 è a^-3/10 = 1 / (a^3/10)

49. Distribute exps, recall 8=2³, so 8^2/3 = (2³)^2/3 = 2² = 4, see book è (32x¹²) / y^16/15 .

50. 

51. see book è x¹⁵ / y^15/2 .

52. Be sure to distribute the exponent to the coefficient.
    

53. 

54. just multiply the exponents out (and move the negatives)

55. see book è (3t^25/6) / s^1/2 .

56. ß move neg exp
    ß simplify some
    ß “divide out” like terms

57. 4/4 = 1 è x à the more correct answer is |x|, so write |x|.

58. 3/3 = 1, 6/3 = 2 è xy² .

59. 3/3 = 1, nothing you can do with the y so leave it alone, see book è

60. 4/2 = 2 è x²y² .

61. Here you take as many ‘whole’ variables out as possible:
    6-5 = 1, 7 – 5 = 2 è

62. Convert these into fractions and it may be easier:
    
    

63. Remember 64 = 2⁶. Again converting these into fractional exponents may make things easier to see:
    ß INCORRECT
    Because x⁶ was the original term we “know” if we take the square root of it we will get positive x³. Thus when we take the cube root of (positive) x³ we must get a positive value. So the correct answer is 2*|x|. Be careful to watch for these things. It is possible even the solutions presented here have made that error.

64. ß INCORRECT
    Again as in 63 notice since we take x⁴ before we take the fourth root – so we “know” x⁴ must be positive and when we take the fourth root we will get a positive number back. (Likewise if we take the square roots of y and z). So the correct answer is è ç.

65 through 68 is just multiplying the numerator and denominator by the stuff on the bottom – effectively you are just multiplying by 1. If you can do one of these you should be able to do all of them.

65. see book

66. a) multiply by
    b) multiply by
    c) multiply by
    
    
67. see book (and problem 68)

68. a) è When not dealing with the square root it may be easier to convert stuff into fractional exponents. Then you multiply by whatever will make the denominator come out as a whole number. For example in this problem the denominator is to the 2/3 power, so multiply by the stuff raised to the 1/3 power as follows:
    è
    
    b) This one the denominator is to the ¾, so multiply by stuff to the ¼:
    
    è
    
    c) This one is trickier as the denominator is to the 4/3 , so multiply 2/3
    (as 4/3 + 2/3 = 6/3 =2, so the exponent will come out as an integer)
    è
    
    

Problems 69 to 74 are just a matter of counting zeros correctly. They are left for you to do.

69. see book è a) 6.93 x 10⁷ b) 2.8536 x 10^-5 c) 1.2954 x 10⁸ .

70. for you to do

71. a) 319,000 b) 0.00000002670 c) 710,000,000,000,000

72. for you to do

73. a) 5.9 x 10¹² mi b) 4 x 10^-13 cm c) 3.3 x 10¹⁹ molecules

74. a) 93 million miles = 93,000,000 miles = 9.3 x 10⁷ miles.
    b) 0.000000000000000000000053 g = 5.3 x 10^-23 g.
    c) 5,970,000,000,000,000,000,000,000 kg = 5.97 x 10²⁴ kg.

75. 7.2*1.806 = 13.0032 – but you had only 2 significant digits in 7.2 so you can have only 2 significant digits in your answer so you say 7.2*1.806 = 13.
    10^-9*10^-12 = 10^-21, so the initial answer is 13 x 10^-21, but we must slide the decimal point one to the left (we must add a tens place)
    so that makes the final answer 1.3 x 10^-20.
    
    (You can check that decimal stuff in your head: see if 13 x 10^-2 = 1.3 x 10^-1?
    They both should come out as 0.13 right?)

76. Notice 8.61 x 10¹⁹ has only 3 significant digits so that is how many your answer will have. So 8.61*1.062 = 9.14382, but we can only take 3 digits so we say that 8.61*1.062 = 9.14. Thankfully this will NOT require any decimal point slide.
    And we have 10²⁴ * 10¹⁹ = 10⁴³, so our final answer is: 9.14 x 10⁴³.

77. See book – both denominator numbers have 4 significant digits è 1.429 x 10¹⁹ .

78. Left for you – the denominator has only 2 significant digits.

79. See book – 0.0058 has only 2 significant digits è 7.4 x 10^-14 .

80. Notice 3.542 x 10^-6 has only 4 significant digits so raising it to the 9^th will also only have 4 significant digits. Likewise the denominator raised to the 12^th will only have 3 significant digits. So our final answer can only have 3 significant digits.
    (3.542 x 10^-6)⁹ = 87747.95658 x 10^-54 à 87750 x 10^-54 à 8.775 x 10^-50 .
    (5.05 x 10⁴)¹² = 275103767.1 x 10⁴⁸ à 275000000 x 10⁴⁸ à 2.75 x 10⁵⁶ .
    
    So we get 8.775 / 2.75 = 3.190909091 and 10^-50 / 10⁵⁶ = 10^-106.
    Thus the final answer is 3.19 x 10^-106.
    
    IMPORTANT – some professors round at each step (as above), others do not (as below). I was taught you are not supposed to round until the very end, however there are many who disagree. Check with your professor.
    
    Without rounding we get:
    8.774795658 / 2.751037671 = 3.189631226 and 10^-50 / 10⁵⁶ = 10^-106.
    Thus the final answer is 3.19 x 10^-106.
    So in this particular problem when you round does not seem to matter (sometimes it will).
    
    

81. Proxima Centauri, the star nearest to our solar system, is 4.3 light years away. Use the information in Exercise 73(a) to express the distance in miles.
    
    Take 4.3 light years * 5.9 x 10¹² miles / light year = 2.5 x 10¹³ miles.
    4.3 * 5.9 x 10¹² = 25.37 x 10¹², slide decimal left and round to 2 significant digits gives you the final answer of 2.5 x 10¹³ miles.

82. The speed of light is about 186,000 miles/sec. Use the information in Exercise 74(a) to find out how long it takes for a light ray from the sun to reach the earth.
    
    9.3 x 10⁷ miles / 1.86 x 10⁵ miles/sec = ?
    9.3 x 10⁷ miles * (1.86)^-1 x 10^-5 sec/mile = ?
    9.3 x 10⁷ miles * (0.537634408) x 10^-5 sec/mile = ?
    
    Your answer should have only 2 significant digits as 9.3 x 10⁷ miles only has 2.
    
    4.973262032 x 10² seconds à 5.0 x 10² seconds should be the final answer.

83. The average ocean depth is 3.7 x 10³ m, and the area of the oceans is
    3.6 x 10¹⁴ m². What is the total volume of the ocean in liters? (One cubic meter of water contains 1000 liters water).
    
    Notice we have only 2 significant digits in either number.
    
    (3.7 x 10³ m) * (3.6 x 10¹⁴ m²) = (3.7*3.6) x (10³ * 10¹⁴) m³ .
    = 13.32 x 10¹⁷ m³ (adjust to proper scientific notation)
    = 1.3 x 10¹⁸ m³ (note decimal shift and exponent increase)
    
    = 1.3 x 10²¹ liters.

84. As of January 2001, the population of the United States was 2.83 x 10⁸, and the national debt was 5.736 x 10¹² dollars. How much was each person’s share of the debt?
    
    Notice 2.83 x 10⁸ has only 3 significant digits.
    
    5.736 x 10¹² dollars / 2.83 x 10⁸ people = (5.736 / 2.83) x (10¹²/10⁸) $/person.
    = 2.026855124 x10⁴ dollars per person. (now adjust to proper scientific notation)
    = 2.03 x 10⁴ dollars per person.
    (notice there was NO decimal shift so the exponent on the 10 stays the same)

85. A sealed room in a hospital, measuring 5 m wide, 10 m long and 3 m high, is filled with pure oxygen. One cubic meter contains 1000 L, and 22.4 L of any gas contains 6.02 x 10²³ molecules (Avogadro’s number). How many molecules of oxygen are there in the room?
    
    First find out how many liters (L) of gas are in the room:
    5m * 10m * 3m = 150 m³.
    150 m³ * 1000 L / m³ = 150,000 L = 1.50 x 10⁵ L.
    
    Now consider that there are 6.02 x 10²³ molecules PER 22.4 L, so we do the
    following (notice how the labels are nicely canceling in all of this):
    Aside: Notice we changed 22.4 L to be 2.24 x 10 L.
    
    
    
    and we “everything” has 3 significant digits so we round and simplify the exponent on the 10 to get the final answer of: 4.03 x 10²⁷ molecules. (Aside: you may ask your professor how the 5 * 10 * 3 comes out to have more than 1 significant digit – I’m assuming they are to be considered exact measurements with infinite precision- not very realistic).
    
    

86. Police use the formula s = (30fd)^1/2 to estimate the speed s (in mi/h) at which a car is traveling if it skids d feet after the brakes are applied suddenly. The number f is the coefficient of friction of the road, which is a measure of the “slipperiness” of the road. The table below gives some typical estimates for f.

	Tar	Concrete	Gravel
Dry	1.0	0.8	0.2
Wet	0.5	0.4	0.1

a) If the car skids 65 feet on wet concrete, how fast was it moving when the brakes were applied.

Here you just look in the table in the concrete column and the wet row and find f = 0.4. You are given that d = 65. So all you need to do is put f and d into the given formula: s = (30fd)^1/2.

= 780^1/2 mi/hr.

And you can probably stop there. However if you really want to simplify further you can do the following:
27.92848009 mi/hr.

b) If a car is traveling at 50 mi/h, how far will it skid on wet tar?

For this one – basically the same deal. However instead of being given d you are given s = 50. Notice also you are now on wet tar so look that up in the table and you find that f = 0.5. Now plug the numbers in and solve for d.

Starting with we get:
As d is under the square root we square both sides:
2500 = 30*0.5*d We now multiply stuff out:
2500 = 15*d And we divide both sides by 15:
2500/15 = d Simplifying we get:
Or rather d » 166.66 feet.

87. Due to the curvature of the earth, the maximum distance D that you can see from the top of a tall building of height h is estimated by the formula:
    D = (2rh + h²)^1/2,
    where r = 3960 miles is the radius of the earth and D and h are also measured in miles. How far can you see from the observation deck of the Toronto CN Tower, 1135 feet above the ground?
    
    First thing – ignore the picture it doesn’t help.
    
    Notice that the height h is given in feet. So you must first convert the height into miles:
    = h.
    
    And now we just plug stuff into the given equation to find D:
    
    So you can see about 41.26192202 miles.

88. Prove the given Laws of exponents for thecase in which m and n are positive integers and m > n.

1. Law 2 (as stated on page 15) is: .
   To prove this we rely on the definition of aⁿ = a*a*a*a…*a
   (i.e. a*a n times), so:
   where the number of a’s on the top = m
   and the number of a’s on the bottom = n.
   
   So if we ‘cancelled’ out all the a’s on the bottom with a’s on the top we would have (m – n) a’s left on the top = a^m-n, since m > n (and both are positive integers).
   
   For a specific example consider if m = 7 and n = 4. Then we would have:
   
   
   
2. Law 5 (as stated on page 15) is:
   Again we will use the notation definition on page 14 to prove this:
   , where appears n >0 times. So we get:
   , where a appears n times in the top
   and b appears n times in the bottom,
   so we can say:
   , and we are done.
   
   
3. Law 6 (as stated on page 17) is:
   To prove this we will use Law 3 which says: a^mn = (a^m)ⁿ.
   And we note that n is positive so –n = -1*n. Thus we can say the following:
   , by the definition of negative exponents we can say that (a/b)^-1 = [1 / (a/b) ] = 1 * (b/a) = b/a. And using that we get:
   

89. Left for you to do.
90. Left for you to do.
91. Left for you to do.
92. Left for you to do.
93. Left for you to do.