Download Scientific Notation

P.2r !ffi
IntegralExponents
and Scientific
Notation
19
Scientific Notation
. bJE
.3E-6
E-3
963BBEBE
?.3E-6
. EEs
* FigureP.l5
Archimedes(287-212 e.c.) was a brilliant Greek inventor and mathematicianwho
studied at the Egyptian city of Alexandria, then the center of the scientific world.
Archimedes used his knowledge of mathematicsto calculatefor King Gelon the
number of grains of sand in the universe: 1 followed by 63 zeros. Of course,
Archimedes'universewas different from our universe,and he performedhis computations with Greek letter numerals,since the modern number systemand scientific
notationhad not yet beeninvented.Although it is impossibleto calculatethe number
ofgrains ofsand in the universe,this story illustrateshow long scientistshavebeen
interestedin quantitiesranging in size from the diameterof our galaxy to the diameter of an atom. Scientific notation offers a convenientway of expressingvery large
or very small numbers.
In scientific notation,a positive number is written as a product of a number bet w e e n I a n d l 0 a n d a p o w e ro f 1 0 . F o r e x a m p l e . 9 . 6 3x 1 0 7a n d 2 . 3 x 1 0 - 6 a r e
numberswritten in scientific notation.
These numbers are shown on a graphing calculator in scientific notation in
ffi
Fig. P.16.Note that only the power of I 0 showsin scientific notationon a calculator.
The graphing calculatorshown in Fig. P.16 convertsto standardnotation when the
ENTER key is pressed(providedthe number is neithertoo large nor too small). tr
Conversionof numbersfrom scientific notationto standardnotation is actuallv
just multiplication.
fuatT/e
p
Scientificnotation to standardnotation
Converteachnumberto standardnotation.
a. 9.63 x 107 b. 2.3 x 10-6
Sotution
a. 9.63x 107: 9.63X 10.000.000Evaltratc
107
: 96,300,000
b. 2.3 x
: -'?? X
"
M o v c d c c i r n l l p o i n l s c v c np l u c c st o t h c r i g h l .
I
1,000,000
FlvaIuatc
: 2.3 x 0.000001
: 0.0000023
7rV 7fu1. Convert
3.78x
M o v c d c c i n r ap
l o i n t s i x p l a c c st o t h c l c l i
to standardnotation.
Observehow the decimalpoint is relocatedin Example 5. Convertingfrom scientific notationto standardnotation is simply a matter of moving the decimalpoint.
To convert a number from scientific to standardnotation, we move the decimal point
the number of placesindicatedby the power of 10. Move the decimal point to the
right for a positive power and to the left for a negative power. Note that in scientific
notationa number greaterthan 10 is written with a positivepower of l0 and a number lessthan 1 is written with a negativepower of 10.NumbersbetweenI and l0 are
not written in scientific notation.
In the next example, we convert from standardnotation to scientific notation by
reversingthe processusedin Example 5.
20
Chapter P r rffi
5. Sel I
Prerequisites
A graphing calculator converts to scientific notation when you press ENTER
ffi
as shown in Fie. P.17. !
6. ESe
fuatT/e
!l Standardnotationto scientificnotation
Converteachnumberto scientificnotation.
" FigureP.17
a. 580,000,000,000
b. 0.0000683
Sotution
a. Determinethe powerof l0 by cor.rnting
the numberof placesthat the decimalmust
movesothatthereis a singlenonzerodigitto theleft ofthe decimalpoint(l 1places).
Since580,000,000,000
is largerthan10,weusea positivepowerof 10:
: 5.8 x l0rl
580,000,000,000
b. Determinethe powerof l0 by countingthe numberof placesthe decimalmust
moveso that thereis a singlenonzerodigit to the left of the decimalpoint (five
places).Since0.0000683
is smallerthanl, we usea negative
powerof l0:
0.0000683:6.83x10-'
77y 77A1. Convert5,480,000to scientificnotation.
One advantageof scientific notation is that the rules of exponentscan be used
when performing certain computationsinvolving scientific notation. Calculators can
be used to perform computations with scientific notation, but it is good to practice
some computation without a calculator.
I
fua*F/a
I
Using scientific notation in computations
Perform the indicated operations without a calculator. Write your answersin scientific notation. Check your answerswith a calculator.
x to-e)r
a.(4x 1or3)(s
fifil,'
c.
(2,000,000,000)'
(0.00009)
600.000.000
Solution
a . ( 4 x 1 0 1 3 ) (x5 l O - e ): 2 0 x l 0 r 3 + ( - e )Productrule for exponents
:20 x l\a
Simplify the exponent.
:2x
Write 20 in scientificnotation
l0rx 104
:2x.105
1 . 2x l 0 - e
4x10-'
b.-10-Y'1'
t.2
4
Productrule for exponents
Quotient rule for exponents
: 0 . 3x 1 0 - 2
Simplify the exponent.
:3X10-3
Use0.3:3 x l0
j.
1
P.2 I Effi
lnteoralExoonentsand SclentificNotation
21
First convert each number to scientific notation, then use the rules of exponents
to simplify:
(0.00009)( 2 x l o e ) 3 ( 9 x t o - 5 )
(2,ooo,ooo,000)'
600,000,000
E
E
( 8 x l o 2 7 ) ( 9x l o 5 )
2eF
1 . ? E- 9 / 4 E- ?
.5E( ? e 9 ) ^ 3 * 9 e- 5 / 6 p
6x108
- 7 2 x l o : 2: D x l o r 4 : 1 . 2x l o r s
6x10"
1 .? e l
* FigureP.18
6X108
These three computations are done on a calculator in Fig. P.18. Set the mode
ffi
to scientific to get the answersin scientific notation.
(7 x 1014)(5
x l0-3).
TrV TTul. Findtheproduct
t
In the next example we use scientific notation to perform the type of computation performed by Archimedes when he attemptedto determine the number of grains
ofsand in the universe.
foo*V/e
I
rne numberof grainsof sand
in Archimedes'eafth
If the radius of the earth is approximately 6.3 8 x I 0r kilometers and the radius of a
grain of sand is approximately I X 10-i meters, then what number of grains of
sand have a volume equal to the volume of the earth?
Solution
Sincethe volume of a sphereis given by V : lrrr3 , thevolume of the earth is
4
i z ( 6 . 3 8 x l o 3 ) 3 k m:3 1 . 0 9x l 0 r 2 k m 3 .
J
Since I km :
is
103m, the radius of a grain of sand is I X 10-6 km and its volume
4
km3- 4.19x lo-18km3.
: r(l x 10-6)3
J
. EH/HH"JHHbE
-l
4, ISSTSB?c
1 . S 9 r l ? r 4 , 1 9 e- l
2.66r43198r
* FigureP.19
To get the number of grains of sand,divide the volume of the earth by the volume of
a grain of sand:
l.0g x 1012km3
4.lg x l0-r8 km3
ffi
- 2.60x l02e
SeeFig.P.l9for thecomputations.
TFV Th.tZ.Findthe volumeof a spherethat hasradius2.4 x l0-3 in.
I
ChapterP rrffi
Prerequisites
For Though
True or False?Explain.Do Not Use a Calculator.
1.2-1 + 2*t :
2 . 2 1 0 0: 4 s 0 ,
lr
3 . 9 8 . 9 8 : 8 1 8r
s.4
5-
t'
27r
4. $.25\-t : 4r
:5-zp
/+\'
:
'
G/
9. l0-a : 0.00001
r
I
6.2.2.2.2-t:
^/:\-'
t.
(a/
--F
IO
10. 98.6X 108: 9.86x 107p
Exercises
M
Evaluate each expression without using a calculator Checkyour
answerswith a calculator (ExampleI)
l. 3-t tlz
2. 2-t tlz
4. -s-3 -tltzs
s. *s
''
I
^-1 - -3-r
1
/q\-:
\r)
st27
/
r\-4
\
J/
t 0 .[ - ; f
3.
-4-2-tlrc
I
v6.
z-
,1
/ o\-2
/
t.(;)
st4
s.
r\-2
\-il
1 3 .l _ " 2 a
o"
a-1
14.
*
rlz+
L5. 20 + 2-r llz
1 7 . * 2 . 1 0 - 3- r l s o o
/rt. -t-, . (-z)-2 -tl+
Simplifueachexpression.(Example
2)
-6xttrtt
19, (-3x2y3)(2xey8)
-ftotobs
20. (-6a7ba)(3a3b5)
2!, x2xa + x3x3 216
22. a1as + a3a9 2at2
39. -1(2x3)2 -+,6
40.(-Zr-r1-r -t
/ -u2\3
41.t
\
"
J
./
R"6
1-;
4 3 . 1 + l .y
\)./
/ . , 2\ a . . s
*'\-;)
h
/ 6xv2 \-3 ra,,
45.l; 41Ll zlx'
\6x y-/
)_,14
'nu.
" ' \ 1_75x__.y",
t8x2y3/ 25y"
29. 52 . 32zzs
30, 23 . 42 tzs
/
.-
-)
o\-)
Simpltfy each expression.Assurnethat all basesare nonzero real
numbers and all exponentsare integers. (Example4)
51.
(rr. ("')(-r) - (-r) - ,(-"') + z(22) +
zl z
/ - r\ - 2
or.\a)
4a2
.t
a\-1 _')<
/1r_\
1a551-4a61zo"
+ m ( 3 m 2 )4 m 3+ m 2
-l
38.(b-\2 - GU-',y+
o
49. (-5az'6-t1t
27. (-m2)(-m) - m(-*)
-r
, /t
gl. (za2)3+ (-3a3)2 na6
24. (4y6)(-xya) + (3xye)(5y) ttxyl}
-38*3
y
'x. -P='q=i
-rpq - L
5p-
- l
'n
,"
1
.
- o m ' n ' z- 1 ^
a
-5il1
47.
26. (w2)(-3v't) - (-5w)(-7w2)
/'
3xy-
23. (-2b\(3b3) + (5b3)(-3b2) -21bs
25. (-2a2)(ae7 -
\3
?34.--:a --^
/
1 6 .6 - r + 5 - t 1 l / 3 0
n. ( !o- 5u) 1ooal3 a')
-9r2r,
617
33. -3x4
zx-
4
1-2
1 2 .4 . 4 , 4 . 4 - t r c
* ( !,- a; ) ( l' - r - u) ],
\2
/\3
/ 6y'
1000
tt.2-1 . 42 . I}-t cls
ar
Simpltfy each expression.Write answerswithout negative exponents.Assume that all variables reoresentnonzero real numbers.
(Example31
/"t
\/r
\ ,
/1
\
-'
P.2 rffim
lnw-zz >3.
notation. (Erantples 5 a.nd(t)
5 6 . 5 . 9 8x 1 0 55 9 8 ^ o o o
5 7 .3 . 5 6x 1 0 - 50 . 0 0 0 0 3 s 6
5 9 . 5 , 0 0 0 , 0 0 0x5 1 0 6
58. 9.333x 10 ',
0.000000009333
10
. 6 s 8r7 l 0 7
60. 16,587,00
61. 0.0000672
6.12/. t0 '
6 2 . 0 . 0 0 0 0 0 0 9e8.18 lx l 0 7
'',
63. 7 x 10 0.000000007
64.6x10-30.006
6 5 . 2 0 , 0 0 0 , 0 0 0 , 0 0x0 l2o ' u
x lo "
66. 0.000000000044
23
abnormal rf yom BMI (from the previous.exercise)satisfies
Convert each number given in standard notation to scientific
notation and each number given in scientific notation to standard
55. 4.3 x 10443,000
Exercises
$hich of the five players does not have abnormal body fat?
Smrth
u Tablefor Exercise
80
8 1 . Displacement-LengthRatio
The displacement-lengthratio D
indicateswhethera sailboatis relativelyheavy or relatively
D is
light (Ted BrewerYacht Design,www.tedbrewer.com).
givenby the formula
Perfbrm the indicated operations without a calculabr. Write your
answersin scientific notation. Use a calculator to check. (li.runtpla7)
6 7 . ( 5 x 1 0 8 ) ( 4x 1 0 1 ) 2 x 1 0 1 6 8 . ( 5 x 1 0 - 1 0 ) ( 6x 1 0 5 )
3 x 1 01
8 . 2x l 0
9 . 3 1 01 2
l0'
I
lo'
69.-2
70'
4.txt0'
t, , ,ut
4 x lo r')
7 1 .5 ( 2 x l o r o ) 3
'''
(2.000.000
)r(0.000005
)
(o,ooo2l2
I x l0rr
u, , :_
2uo
'
where I is the length at the water line in feet, and d is the displacementin pounds.IfD > 300, then a boat is considered
relativelyheavy.Find D for the USS Constitution,the nation's
oldest floating ship, which has a displacementof 2200 tons
and a length of 175 feet. 1J- 36(r.5
i 2 . 2 ( 2 x 1 0 8 ) - 4l . 2 s x l o r r
t+'
dLt.t06
-3
(6,ooo,ooo)2
(o.ooooo3)
( I ,000,000)
(2000)3
1 . 7x 1 0 1 1
82. A Lighter Boat IfD < 150 (frorn the previousexercise),
Use a calculator to perfbrm the indicated operations. Give your
answersin scientific notation. (lixumplt'8)
/5.
then a boat is consideredrelatively light.
a. Use the accompanyinggraph to estimatethe length for
which D < 150for a boat with a displacement
of I 1,500
pounds.Lcngthgrcatcr'1han
l0 11
( 4 . 3 2> <l 0 - e ) ( 2 . 3x 1 0 4o) . o 3 rx, r o 5
7 6 . ( 2 . 3 3> <1 0 2 3 ) ( 3 . x9 8l 0 - e )( ) . 2 7 3x4 t 0 t 4
b. Use the formula to find D for a 50-foot boat with a displacementof I 1,500pounds.4 l. I
( 5 . 6 3x t o - 6 ) 3 ( 3 .x5 l 0 7 ) 4
4 . 7 8x l 0 " '
77.
x 10 4)2
?7'.(8.9
78.
rQ.39 x l0-r2)2
sl{'r lo
\ 6 . 7 5/ t o \
gt""
riii
t i l rf l
;rti'-c400
rli
b4
ii,i :oo
Solve each problem. Useyour calculator.
79. Body-Mass Index The body mass index (BMI) is used to assessthe relativeamountof fat in a person'sbody (ShapeUp
America, www.shapeup.org).If w is weight in poundsand /l is
height in inches,then
r;rE 200
rli i ra
.9 100
rli
.i,
li;'
,I
BMI : 703wh-2.
Find the BMI for Dalias Cowboys'playerFlozell Adams at
6'7" and335 pounds.BMI :37.7
80. Abnormal Fat The accompanyingtable showsthe height and
weight of five Da1lasCowboys players in training camp (ESPN,
www.espn.com).Somephysiciansconsideryour body fat
o--
25
50
]s
Length (f'eet)
, i l , i I i , , t r f ti : , 1 ; i l l l l i | i l , 1 , ' l : t . i iIir f i . r i
m Figure for Exercise 82
83. Nalional Debt In February2005,the nationaldebt was
1.634 x 1012dollarsandthe U.S.populationwas 2.956 X 108
people(U.S.Treasury,www.treas.gov).What was eachperson'sshareof the debt at that time? $25.825
24
C h a p t e r Pr s s b
Prerequisites
84. IncreasingDeh
In the three yearsprior to Februaryof 2005,
the national debt was increasingat an averagerate of L44 X 10"
dollars per day. If the debt keeps increasing at that rate, then
what will be the amount of the debt in February of 20 11? See
s x e r c i s e$.1 . 0 7 9 l 0 l '
t h ep r e v i o u e
85. Energyfrom the Sun Merely an averagestar, our sun is a
swirling mass of densegasespowered by thermonuclearreactions.The great solarfurnacetransforms5 million tons of
mass into energy every second.How many tons of masswill
be transformed into energy during the sun's 10 billion-year
iifetime? 1.577x l02atons
F i n dt h e n u m b e o
r f s e c o n disn l 0 b i l l i o ny e a r s .
89. Mass oJ the Sun The massof the sun is 1.989 x 1030kg and
the massof the earth is 5.976 r 1024kg. How many times
larger in massis the sun than the earth?3.3 x 105
90. Speedo/'Light Ifthe speedoflight is 3 X 108m/sec, then
how long does it take light from the sun to reach the earth?
(SeeExercise88.) 8.-lrnin
'iiiii!,ft r -- DIR
For Writing/Discussion
91. The integral powers of 10 are important in the decimal number
system.
a. Use a calculatorto evaluate5365.4
5 . 1 0 r + 3 . 1 0 2+ 6 . l 0 r + 5 ' 1 0 0+ 4 . l 0 - r .
b. Without using a calculator,evaluate9102.38
9 . 1 0 3+ 7 . 1 0 2+ 2 . r c o + 3 . l 0 - r + 8 . 1 0 * 2 .
c. Write a brief explanationof how you did part (b).
n Figurefor Exercise
85
86. Orbit of'the Earth The earthorbits the sun in an approximatelycircularorbit with a radiusof 1.495979X 108km.
What is the areaof the circle?7.03x l0r('krn2
d. Write 9063.241as a sum of integralmultiples of powersof
10. Is your answerunique?Write a detaileddescriptionof
h o w y o u d i d i t . g x l 0 r + ( r X 1 0 1+ 3 x 1 0 0 + 2 x 1 0 1 +
4x l0 2+ I x l0 r,notuniquc
e. Try to write 43,002.l9 as a sum of integralmultiples of
powersof l0 using the descriptionthat you wrote for
part (d).
4 x 1 0 1 + 3 x l 0 r + 2 x 1 0 0 +I x l 0 I + . ) x l 0 2
92. Many calculatorscan handlescientific notationfor powersof
l0 between-99 and 99. How can you be expectedto compute
(4 X l}2201zlUsctlrcrulcsofcxponents
to gct l.(r x 1014r.
fta,qs 0*ft1!e
!h,eeu !! { !t!
Powers oJ'Three
What is x if
x 108km
1.495979
1
27
' Figurefor Exercises
86 and 87
8 7 . Radiusof the Earth The radiusof the sun is 6.9599 / 10s
km, which is 109.1times the earth'sradius.What is the radius ofthe earth?6379km
88. Distance to the Sun The distancefrom the sun to the earth is
| .495979 X 108km. Use the fact that I km : 0.621 mr ro
find the distancein miles from the earth to the sun.
9 . 2 9X 1 0 7m i
3 1 0 0.
CommonRemainders
The numbers
|,576,231,
4,080,602,
and
2,690,422
all yield the sameremainderwhen divided by a certainnatural
number that is greaterthan one. What is the divisor and the
remainder'?Divisor89.rcmaindcr'
41