Solutions to the Even-Numbered Exercises in Chapter 4
Exercises for Section 4.1
2. a. 0.000 000 1
b. 10,000,000
c. –100 000 000
d. –0.000 01
4. a. 1 millisecond
6. a. 1013
b. 10-12
e. 0.001
f. 100,000
b. 1 kilogram
c. 104
d. 10-4
c. 1 hectometer
e. 106
f. 10-6
8. Each violates the rule that in scientific notation the number is written as N • 10n,
where 1 ≤ |N| < 10. Rewriting each in scientific notation, we have:
a. 2.5•105
b. 5.6•10-4
c. 1.2•10-4
d. –4.2503•104
10. a. 1.37•1010 years
b. 5.0•10-5 meters
c. 6.4•103 km
d. 2.0•108 years
e. 5.0•10-5 meters (same as 10. b.)
12. a. 1.67•10-27 grams
b. 1.6•10-5 m3
14. a. 8
c. 4
16. a. –7 and 7
b. 6 and –4
b. 4
c. 9 and – 5
d. none
c. 5•10-8 m
d. 1
e. 9 and –5
f. 4 and -4
18. a. (0, 0) and (2, 35)
b. 35/2 = 17.5
c. It means that for each additional billion of light years from Earth, the recession
velocity increases by 17.5 thousand kilometers per second.
d. y = 17.5x
20. Note: Since the coordinates given below come from eyeball estimates, student
answers may vary from those given here.
a. Wolcott (0.1, 1900) i.e, (0.1, 1900) ; Sollas (0.2,1908); and Clarke (0.3, 1921).
b. Barrell did it in 1918; the coordinates are (1.3, 1918).
c. Estimating two points on a hand—drawn line could result in coordinates such as (0,
1910) and (4, 1956). Then the slope, m, of the line is
m= (1956 – 1910) / (4 – 0) = 46/4 = 11.5
A plot of the various estimates and the graph with this slope and y-intercept are
given at the top of the next page on the left
d. One meaning for the slope is that for each increase of a billion years in the
estimated age of Earth, time advances, on average, 11.5 years. However it would
make more sense to say that on average for every 11.5 years, estimates of Earth’s
age increased by a billion years.
59
Solutions to the Even Numbered Exercises in Chapter 4
1970
Tilton
1960
•
•
Patterson
Holmes
•
1950
1940
•
Polkanov
Ellsworth
•
1930
1920
Clarke
•
1910 •
• Barrell
Sollas
Wolcott
1900 •
1
2
3
4
5
Billions of years
6
Exercises for Section 4.2
c. a12
d. -8a6
22. a. 1
b. -1
e. 8a12
f. -8a12
g. 1000a6 b9
h. a2 b2
3 2
9


24. a.    
5 
25
3 4
5a 
b.  2   54 a 4  625a 4
 a 
10a 3 
102 a 6 100a 6 4a6
c.
 2

  2 2 
 5b 
5 b
25b2
b
3 3
3 9
9
2x  2 x
8x
d.  2   3 6 
 3y 
3 y
27y 6
26. a. -16 + 4 = -12
b. -8 + 16 = 8
c. 18 - 12 = 6
d. 100,000 – 10,000 = 90,000
e. 1,000 + 8 = 1008
f. 2,000+1,000+100 = 3,100
g. 2,000 – 1,000 = 1,000
h. 1
10
10
11
28. a. 6•10 + 4•10 = 10
30. a.
   x   x
3 5
x 2 
6 5
2
2 6 • 10 24
 1018
b. 6
6
2 • 10
c. 2•106•4•104 = 8•1010.
30

7x 2 y 6
4

7y
x2y2
4 x 3 y 5 3 64 x 9 y15
8 6 3
c. 


x y
4 
3 12
27
 6xy  216x y
d. 1.21 • 1010
e. 0.5 • 103 = 5 • 102
f. 6 • 103 + 7 • 105 = 0.06 • 105 + 7 • 105 = 7.06 • 105
b.



32. a. Population of China = 1.284•109; population of Monaco = 3.199•104.
Area of China = 3.704•106 miles2; area of Monaco = 7.5•10-1 miles2
60
Solutions to the Even-Numbered Exercises in Chapter 4
b. (1.284•109)/(3.199•104) = 4.014•104. So China’s population is 4 orders of
magnitude larger than Monaco’s..
c. China's population density = (1.284•109)/( 3.704•106) 3.467•102  350
people/mile2. Monaco's population density = (3.199•104 )/(7.5•10-1) = 4.265•104 =
42,650 people /mile2.
d. Student answers will vary but they should mention that while China's area and
population are much bigger than Monaco's, Monaco has a larger population
density; which is 2 orders of magnitude larger than China’s..
34. 1.4112 • 1021 miles
36. a. (6•105)/(3•102) = 2•103 or $2000 per person..
b. (2.5•109)/(5•102) = 5•106 or 5 million watts per hour.
c. (6•106)/(3•10) = 2•105 or 200,000 births per year.
38. a. 2,968,001,000•189,000  3•109• 2•105 = 6•1014 (compared to 5.6095 •1014 using a
calculator.
b. 0.000079•31,140,284,788  8•10-5 •3•1010 = 2.4•106 (compared to 2.4601•106
using a calculator).
c. (4,083,693•49,312)/(213•1945)  (4•106•5•104 )/(2•102•2•103) = 5•105 (compared
to 4.8608•105 using a calculator).
40. Using Rule 3: a 2   a 2•3  a 6 and a 3   a 3•2  a 6 . Students may use other laws.
3
2
42. (2.5•109 ) ( 3•109 ) = 7.5•1018 bacteria per year.
44

nfactors of a
a
(a •

m
a
a •
n
If n >m > 0 then
• a)
 anm since each of the m a's on the bottom
• a
m factors of a
cancels with an a on top and there are n - m a's still left on top.
n factors of a

a
(a •
If n = m, then n 
a
a •
n
• a)
1  a0  ann
• a
n factors of a
Exercises for Section 4.3

46. a. x-3+4 = x1
b. x-3-2 = x-5 = 1/x5
c. x(2•-3) = x-6 = 1/x6
d. n(-2•-3) = n6
e. 2-3•n(-2•-3) = n6/23
f. n - n-2 + n-2 - n = 01
61
Solutions to the Even Numbered Exercises in Chapter 4
48. a. 3-18 = 1/318
b. x11
1
c. 2 8  4
2
d. 2x-3 + 3x-3 = 5/x3
e. no simpler form
50. a. 7.25 • 1025
b. 7.25 • 10-21
c. 1.3793 • 10-26

d. -7.25 • 1025
e. -7.25 • 10-21
20
2 • 10

 104 =0.000 1
5
200,000 2 • 10
0.006 6 •103
b.

107 = 0.000 0001
4
60,000 6 •10
c. 200 • 0.000 007 5 = (2•102 ) • (7.5•10-6 )= 1.5•10-3 = 0.0015
10,000,000
107
d.

 4 •102 = 400
4
25,000
2.5 •10
e. 0.06 • 600 = (6•10-2) • (6•102 )= 3.6•10=36
f. 10% of 0.000 05 = (1•10-1) • (5•10-5 ) = 5 •10-6 = 0.000 005
52. a.


 54. a. -4
b. 0.001
c. –3
d. 0.1
e. -11
f. –8
56. a. multiplies instead of adding exponents; x-7
b. transfers variables with negative exponents from top to bottom and vice versa
21 x 2 y 2
x2x
x3


without changing the sign of the variable's exponent;
x 1 y 5
2y 2 y 5 2y 7
2
9
c. puts 3 in denominator with x; 3 • x 1  2
x
1
d. assumes that the inverse of a sum is the sum 
of inverses; (x + y)-1 =
xy
--4
-6
-10
-10
58. a. (0.000 359)•(0.000 002
 47)  4 •10 •2 •10 = 8 •10 (compared to 8.867•10
using a calculator)
b. (0.000 007 31•82,560)/1,891,000  (7•10-6•8•104)/(2•106) = 2.8•10--7 (compared to

3.19•10-7 using a calculator).
60. a. 5.803•1012
b. 3.436•1010
c. 4.572•10-4
62.
d. 1.489•10-11
e. 1.049•10-4
f. –3.638•10-12
5
7
95
 9 5 • 27 7  32  33   310 • 321  331
7
27
64. a. (500 dollars)  (20 nickels/dollar) = 10,000 nickels = 104 nickels.
C 40.2 • 10 6 meters
1 km
40.2 • 10 3 km

• 3

 6.398 • 10 3 km
b. r 
2
2
10 meters
2


62
Solutions to the Even-Numbered Exercises in Chapter 4
66. a. 160,000,000/186,000 = 860.2 seconds or 14.3 minutes
b. 45•106/3•105 = 150 seconds
Exercises for Section 4.4
1 mile
15.538 miles
1.609 km
1 ft
1 yard
b. 700 m •
•
 765.027 yards
0.305 m 3 ft
1 inch

c. 250 cm •
 98.425 in.
2.54 cm
68. a. 25 km •


1 oz
1.764 oz
28.35 gr
1 lb
e. 10 kg•
 22.046 lbs.
0.4536 kg
1 qt
f. 10000 ml •
10.571 qts.
946 ml
d. (50 gr) •

3 ft 1 meter
•
 91.463 m  0.0915 km
1 yd 3.28 ft


72. a. The conversion factor to go from feet to meters is 0.305 m / 1 foot. So the
conversion factor from square feet to square meters is (0.305m)2/(1 foot)2 = 0.093
m 2 / 1 ft 2 . So 50 ft 2 = 50 ft 2  (0.093 m 2)/(1 ft 2 ) = 4.65 m 2

2
2 100 cm
b.. Thus the volume of roll of foil = 4.65 m •
• 0.0038 cm  176.7 cm 3
1 m2
2.7 gr
and thus the weight of the roll of foil is
• (176.7 cm 3 )  477 grams.
1 cm 3
1 ft
1 ft
74. a. (1012 m) •
• 1012 ft
b. (1015 m) •
3.2787
 3.279 • 10-15 ft
0.305 m
0.305 m

70. 100 yd •

76. a. 1.5 • 108 km and 1.08 • 108 km.
-1
b. 7.2 • 10
c. 0.72 A.U.
d. 39.3 A.U.

78. a. Circumference of Earth’s orbit = 2π•9.3•107 = 5.843 •108 miles.
b. 365 days •24 hours / day = 8760 hours [Actually, the year is 365.25 days and thus
it is really 8766 hours .]
c. Orbital speed = (5.843 •108 miles)/(8760 hr)  66,700 mph.
4.0833 ft 3
 1.24 ft 3 per bu
80. a.
3.2812 bu
(12 in) 3
 7055.94 in 3
b. 4.0833 ft 3 •
1 ft 3
(30.5 cm) 3

3
1.244
ft
•
 35295.55 cm 3
c.
3
1 ft

82. Student answers will differ because of different estimates of B = the number of heart
beats per minute and of Y = number of years in one's lifetime. Now there are 525,600

minutes in a year and thus the total number of heart beats in one's lifetime is
525600•B•Y. Taking B = 70 and Y = 75 we get a total = 2.7594•109 heart or almost 3
billion beats in a lifetime.
63
Solutions to the Even Numbered Exercises in Chapter 4
84. a. 1600 years
b. 200 years apart; one took 300 years to get to Earth and the other 500 years and they
have arrived at the same time.
86. Volume of flower bed mulch = [(25 • 3) + (15 • 4) + (30 •1.5)](4/12) = 60 cu. ft. and
there are 27 cu. ft. in a cubic yard, so one needs 60/27 = 2.22 cu. yd of mulch. So one
must buy 3 cu. yd. at $27 a cu. yd. for a total of $81.
88. a. 1 angstrom = 10-10 meters = 10-8 cm. Thus 3900 angstroms = 3.9•103•10-8 =
3.9•10-5 cm and 7700 angstroms = 7.7•10-5 cm.
b. 0.0001 angstroms = 10-4•10-8 = 10-12 cm.
c. 1 nm = 10 angstroms = 10•10-8 cm = 10-7 cm. = 10-9 m.
90. a. 1 megabyte = 1024 kilobytes and 1024 = 210 = 1.024•103
b. 1 gigabyte = 230•23 or 233 bits and 233 = 8,589,934,592  8.59,•109 or well over 8
billion bits
Exercises for Section 4.5
92. a. 10000  [10 4 ]0.5  10 2  100
b. 25 is undefined as a real number.
c. 6250.5 = 625 = 25
d. 1000.5 = 10
e. (1/9)0.5 = 1/3
f. (625/100)0.5 = 25/10 = 2.5
94. a. 2 50 12 8  2 25 • 2 12 4 • 2 10 2  24 2  34 2
b. 3 27  2 75  3 9 • 3  2 25 • 3  9 3 10 3   3
c. 10 32  6 18 10 16 • 2  6 9 • 2  40 2 18 2  22 2
d. 2 3 16  4 3 54  2 3 8 • 2  4 3 27 • 2  4 3 2 12 3 2 16 3 2


35
 3.338 ft
 96. r =


98. a. 2
e. 4
b. –2
f. -4
c. 3
g. 2
d. -3
h. 8
100. s  3 6  1.817 cm
102. a. True
b. True
c. True
104. a. 64 3  4
b.  144  12
1
c.



81x
1
4
  (81x)
3
d.
3
4
 27x

3
4
5
243  3
e. 4 165  32

64
d. False
Solutions to the Even-Numbered Exercises in Chapter 4
106. a.
13
is between 3 and 4 since 9 < 13 < 16.
b. radical 22 is between 4 and 5 since 16 < 22 < 25.
c. 40 is between 6 and 7 since 36 < 40 < 49
108. Constellation: The simplified powers ua•mb are:
y
20
a.
b.
c.
d.
u2 • m5
u-24/5 • m3
u-5 • m6
u1 • m2
e. u-7 • m8
f. u-12 • m9
g. u-9 • m37/4
Constellation
The graph on the right of the points (a, b)
in Constellation looks like the Big Dipper
(Ursa Major).
0
-20
x
10
110. Surface area = area of base + 4•(area of 1 triangle)
= (30•30 )+ (4•0.5•30)• 900  225  2460 sq. in.
112. ramp length =
4 2  48 2  2320  48.2 ft.
114. a. 31.342
b. 70.166 
c. 1762.496
Exercises 
for Section 4.6
116. a. reduces; 6
b. increases; 3
c. 48
118. 1 /10-3 = 1•103 or 3 orders of magnitude taller
120. Temperature of core of the sun / temperature of boiling water = 2.0•107
kelvins/(3.73•102)kelvins = 5.36•104 So the sun’s temperature is 4 orders of
magnitude hotter than the temperature at which water boils at sea level.
122. 23 spores since 23000/23 = 103.
124. a. The solar system is 5 orders of magnitude larger than Earth.
b. A proton is 36 orders of magnitude smaller than the Milky Way.
c. An atom is 5 orders of magnitude larger than a neutron.
126. a. 10-3/10-5=102 ; thus rain water is 100 times more acidic than preindustrial rain.
b. 10-3.4/10-7.4 = 104 ; thus wine is 4 orders of magnitude more acidic than blood.
128. a. 1 meter = 100 meters
b. 10 meters = 101 meter.
c. 1 hectometer = 100 meters = 102 meter.
(Plot on the logarithmic scale is omitted.)
65
d. 1000 kilometers = 106 meters
e. 10 gigameters = 1010 meters
.
Solutions to the Even Numbered Exercises in Chapter 4
130. a. No answer necessary.
b. One way to furnish some common ground for all student answers is to use the
following mind set:
In Sagan's table the gamut of cosmic events is condensed to a time line for one
year. The student is here asked to relate the gamut of these events to his/her time
span. This latter could be the number of years the student has been alive. In order
to make the transition from Sagan's time span to the student's one need only
compute the percentage of time that went by in the condensed year from the
beginning to each cosmic event and then apply that percentage to the student's
time span.
Below is a table that gives such percentages for the particular cosmic events asked
about.
Cosmic Ev ent
Big Bang
Creation of Earth
First Life on Earth
First Homo Sapiens
American Revolution
Sagan Date
In 1 yr. condensation
Jan. 01
Sep. 14
Sep. 25
Dec. 31 10:30 PM
Dec. 31 11:59 PM
Percentage of
that year
00.00000
70.41096
73.42466
99.98288
99.99981
Note: The Sagan dates in the one year condensation for the American Revolution
are estimates based on other estimates found in the reading. Students may give a
different one, e.g., midnight or slightly thereafter. Keep in mind that requested
student paragraph is meant to be a playful, imaginative exercise.
132. a. "Mega" means a multiplying factor of 106. Thus 104 MHz = 1.04•108 Hz.
b. "Kilo" means a multiplying factor of 103. Thus 1030 kHz = 1.030•106 Hz.
c.  = (3•108)/(1.04•108)  2.89 meters.
d.  = (3•108)/1.03•106)  2.91•102 or 291meters.
e. A football field is 2 orders of magnitude larger than the FM wavelength given in
part c. and of the same order of magnitude as the AM wavelength found in part d.
Exercises for Section 4.7
134. a. 2.51
b. 3.16
c. 3.98
d. 5.01
e. 6.31
f. 7.94
136. a. 1000 < 4000 < 10000  3 = log(1000) < log(4000) < log(10000) = 4 
log(4000) is between 3 and 4.
b. 1,000,000 < 5,000,000 < 10,000,000  6 = log(1,000,000) < log(5,000,000) <
log(10,000,000) = 7  log 5, 000,000 is between 6 and 7.
c. 0.0001 < 0.0008 < 0.001  –4 = log(0.0001) < log(0.0008) < log(0.001)= -3 
log(0.0008) is between -4 and –3.
66
Solutions to the Even-Numbered Exercises in Chapter 4
138. a. i. 2 ii. 3 iii. 7. The logarithm values are increasing.
b. i. –1 ii. –3 iii. -5. The magnitude of the log values are getting larger but
remain negative.
c. It is not defined.
d. log(-10) is not defined. The log function is defined for positive numbers only.
140. a. x = 101.079 12
b. x = 100.699 5
c. x = 102.1 126
d. x = 103.1  1259
142. a. x = log (153)  2.185
b. x = log (153000)  5.185
c. x = log( 0.125)  -0.903
d. x = log(0.00125)  -2.903
144. a. 10 x – 2 = 102  x –2 = 2  x = 4
b. log(x – 4) = 1  x – 4 = 10  x = 14
146. a. 0.01
b. 0.001
c. 2x – 3 = 3  x = 3
d. 6 – x = 0.01  x = 5.99
c. 0.0001
148. a. log 15 ≈ 1.1761: Check 101.1761  15
b. log 15,000 ≈ 4.1761; Check 10 4.1761  15,000
c. log 1.5 ≈ 0.1761; Check 100.1761  1.5
150. a. 10•log(10-8/10-15) = 10•log(107) = 70 decibels
b. log 50 = 1.699  50 = 10 1.699. So 50 would lie off-center, between 10 1.6 and
10 1.7 on the multiplicative scale.
152. a. 50 goes on the middle mark in the additive scale.
b. log 50 = 1.699 and thus would go between 101.6 and 101.7 and very close to 101.7.
The actual placement on the log scale is omitted here.
154. a. 125 = 102.097 b. 372 = 102.571
= 102.924
c. 694 = 102.841
d. 840
The plots of these points on the log scale are omitted. Each point would be
proportionally placed between the nearest integer powers of 10 on the given scale.
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67