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Scientific Notation

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary: This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The basic operations with real numbers are presented in this chapter. The concept of absolute value is discussed both geometrically and symbolically. The geometric presentation offers a visual understanding of the meaning of |x|. The symbolic presentation includes a literal explanation of how to use the definition. Negative exponents are developed, using reciprocals and the rules of exponents the student has already learned. Scientific notation is also included, using unique and real-life examples. Objectives of this module: be able to convert a number from standard form to scientific form and from scientific form to standard form, be able to work with numbers in scientific notation.

Overview

  • Standard Form to Scientific Form
  • Scientific Form to Standard Form
  • Working with Numbers in Scientific Notation

Standard Form to Scientific Form

Very large numbers such as 43,000,000,000,000,000,000 (the number of different possible configurations of Rubik’s cube) and very small numbers such as 0.000000000000000000000340 0.000000000000000000000340 (the mass of the amino acid tryptophan) are extremely inconvenient to write and read. Such numbers can be expressed more conveniently by writing them as part of a power of 10.

To see how this is done, let us start with a somewhat smaller number such as 2480. Notice that

2480 Standardform =248.0× 10 1 =24.80× 10 2 = 2.480× 10 3 Scientificform 2480 Standardform =248.0× 10 1 =24.80× 10 2 = 2.480× 10 3 Scientificform

Scientific Form

The last form is called the scientific form of the number. There is one nonzero digit to the left of the decimal point and the absolute value of the exponent on 10 records the number of places the original decimal point was moved to the left.

0.00059 = 0.0059 10 = 0.0059 10 1 = 0.0059× 10 1 = 0.059 100 = 0.059 10 2 = 0.059× 10 2 = 0.59 1000 = 0.59 10 3 = 0.59× 10 3 = 5.9 10,000 = 5.9 10 4 = 5.9× 10 4 0.00059 = 0.0059 10 = 0.0059 10 1 = 0.0059× 10 1 = 0.059 100 = 0.059 10 2 = 0.059× 10 2 = 0.59 1000 = 0.59 10 3 = 0.59× 10 3 = 5.9 10,000 = 5.9 10 4 = 5.9× 10 4

There is one nonzero digit to the left of the decimal point and the absolute value of the exponent of 10 records the number of places the original decimal point was moved to the right.

Scientific Notation

Numbers written in scientific form are also said to be written using scientific notation. In scientific notation, a number is written as the product of a number between and including 1 and 10 (1isincluded,10isnot) (1isincluded,10isnot) and some power of 10.

Writing a Number in Scientific Notation

To write a number in scientific notation:

  1. Move the decimal point so that there is one nonzero digit to its left.
  2. Multiply the result by a power of 10 using an exponent whose absolute value is the number of places the decimal point was moved. Make the exponent positive if the decimal point was moved to the left and negative if the decimal point was moved to the right.

Sample Set A

Write the numbers in scientific notation.

Example 1

981

The number 981 is actually 981. 981. , and it is followed by a decimal point. In integers, the decimal point at the end is usually omitted.

981=981.=9.81× 10 2 981=981.=9.81× 10 2

The decimal point is now two places to the left of its original position, and the power of 10 is 2.

Example 2

54.066=5.4066× 10 1 =5.4066×10 54.066=5.4066× 10 1 =5.4066×10

The decimal point is one place to the left of its original position, and the power of 10 is 1.

Example 3

0.000000000004632=4.632× 10 12 0.000000000004632=4.632× 10 12

The decimal point is twelve places to the right of its original position, and the power of 10 is 12 12 .

Example 4

0.027=2.7× 10 2 0.027=2.7× 10 2

The decimal point is two places to the right of its original position, and the power of 10 is 2 2 .

Practice Set A

Write the following numbers in scientific notation.

Exercise 1

346

Solution

3.46× 10 2 3.46× 10 2

Exercise 2

72.33 72.33

Solution

7.233×10 7.233×10

Exercise 3

5387.7965 5387.7965

Solution

5.3877965× 10 3 5.3877965× 10 3

Exercise 4

87,000,000

Solution

8.7× 10 7 8.7× 10 7

Exercise 5

179,000,000,000,000,000,000

Solution

1.79× 10 20 1.79× 10 20

Exercise 6

100,000

Solution

1.0× 10 5 1.0× 10 5

Exercise 7

1,000,000

Solution

1.0× 10 6 1.0× 10 6

Exercise 8

0.0086 0.0086

Solution

8.6× 10 3 8.6× 10 3

Exercise 9

0.000098001 0.000098001

Solution

9.8001× 10 5 9.8001× 10 5

Exercise 10

0.000000000000000054 0.000000000000000054

Solution

5.4× 10 17 5.4× 10 17

Exercise 11

0.0000001 0.0000001

Solution

1.0× 10 7 1.0× 10 7

Exercise 12

0.00000001 0.00000001

Solution

1.0× 10 8 1.0× 10 8

Scientific Form to Standard Form

A number written in scientific notation can be converted to standard form by reversing the process shown in Sample Set A.

Converting from Scientific Notation

To convert a number written in scientific notation to a number in standard form, move the decimal point the number of places prescribed by the exponent on the 10.

Positive Exponent Negative Exponent

Move the decimal point to the right when you have a positive exponent, and move the decimal point to the left when you have a negative exponent.

Sample Set B

Example 5

4.673× 10 4 4.673× 10 4 .

The exponent of 10 is 4 so we must move the decimal point to the right 4 places (adding 0's 0's if necessary).

Number written in scientific notation as 'four point six seven three zero multiplied by ten to the fourth power' is equal to 'forty six thousand seven hundred thirty' in standard form.

Example 6

2.9× 10 7 2.9× 10 7 .

The exponent of 10 is 7 so we must move the decimal point to the right 7 places (adding 0's 0's if necessary).

2.9× 10 7 =29000000 2.9× 10 7 =29000000

Example 7

1× 10 27 1× 10 27 .

The exponent of 10 is 27 so we must move the decimal point to the right 27 places (adding 0's 0's without a doubt).

1× 10 27 =1,000,000,000,000,000,000,000,000,000 1× 10 27 =1,000,000,000,000,000,000,000,000,000

Example 8

4.21× 10 5 4.21× 10 5 .

The exponent of 10 is 5 5 so we must move the decimal point to the left 5 places (adding 0's 0's if necessary).

4.21× 10 5 =0.0000421 4.21× 10 5 =0.0000421

Example 9

1.006× 10 18 1.006× 10 18 .

The exponent of 10 is 18 18 so we must move the decimal point to the left 18 places (adding 0's 0's if necessary).

1.006× 10 18 =0.000000000000000001006 1.006× 10 18 =0.000000000000000001006

Practice Set B

Convert the following numbers to standard form.

Exercise 13

9.25× 10 2 9.25× 10 2

Solution

925

Exercise 14

4.01× 10 5 4.01× 10 5

Solution

401000

Exercise 15

1.2× 10 1 1.2× 10 1

Solution

0.12 0.12

Exercise 16

8.88× 10 5 8.88× 10 5

Solution

0.0000888 0.0000888

Working with Numbers in Scientific Notation

Multiplying Numbers Using Scientific Notation

There are many occasions (particularly in the sciences) when it is necessary to find the product of two numbers written in scientific notation. This is accomplished by using two of the basic rules of algebra.

Suppose we wish to find (a× 10 n )(b× 10 m ) (a× 10 n )(b× 10 m ) . Since the only operation is multiplication, we can use the commutative property of multiplication to rearrange the numbers.

(a× 10 n )(b× 10 m )=(a×b)( 10 n × 10 m ) (a× 10 n )(b× 10 m )=(a×b)( 10 n × 10 m )

Then, by the rules of exponents, 10 n × 10 m = 10 n+m 10 n × 10 m = 10 n+m . Thus,

(a× 10 n )(b× 10 m )=(a×b)× 10 n+m (a× 10 n )(b× 10 m )=(a×b)× 10 n+m

The product of (a×b) (a×b) may not be between 1 and 10, so (a×b)× 10 n+m (a×b)× 10 n+m may not be in scientific form. The decimal point in (a×b) (a×b) may have to be moved. An example of this situation is in Sample Set C, problem 2.

Sample Set C

Example 10

(2× 10 3 )(4× 10 8 ) =(2×4)( 10 3 × 10 8 ) =8× 10 3+8 =8× 10 11 (2× 10 3 )(4× 10 8 ) =(2×4)( 10 3 × 10 8 ) =8× 10 3+8 =8× 10 11

Example 11

(5× 10 17 )(8.1× 10 22 ) =(5×8.1)( 10 17 × 10 22 ) =40.5× 10 1722 =40.5× 10 5 (5× 10 17 )(8.1× 10 22 ) =(5×8.1)( 10 17 × 10 22 ) =40.5× 10 1722 =40.5× 10 5

We need to move the decimal point one place to the left to put this number in scientific notation.

Thus, we must also change the exponent of 10.

40.5× 10 5 4.05× 10 1 × 10 5 4.05×( 10 1 × 10 5 ) 4.05×( 10 15 ) 4.05× 10 4 40.5× 10 5 4.05× 10 1 × 10 5 4.05×( 10 1 × 10 5 ) 4.05×( 10 15 ) 4.05× 10 4

Thus,

(5× 10 17 )(8.1× 10 22 )=4.05× 10 4 (5× 10 17 )(8.1× 10 22 )=4.05× 10 4

Practice Set C

Perform each multiplication.

Exercise 17

(3× 10 5 )(2× 10 12 ) (3× 10 5 )(2× 10 12 )

Solution

6× 10 17 6× 10 17

Exercise 18

(1× 10 4 )(6× 10 24 ) (1× 10 4 )(6× 10 24 )

Solution

6× 10 20 6× 10 20

Exercise 19

(5× 10 18 )(3× 10 6 ) (5× 10 18 )(3× 10 6 )

Solution

1.5× 10 25 1.5× 10 25

Exercise 20

(2.1× 10 9 )(3× 10 11 ) (2.1× 10 9 )(3× 10 11 )

Solution

6.3× 10 20 6.3× 10 20

Exercises

Convert the numbers used in the following problems to scientific notation.

Exercise 21

Mount Kilimanjaro is the highest mountain in Africa. It is 5890 meters high.

Solution

5.89× 10 3 5.89× 10 3

Exercise 22

The planet Mars is about 222,900,000,000 meters from the sun.

Exercise 23

There is an irregularly shaped galaxy, named NGC 4449, that is about 250,000,000,000,000,000,000,000 meters from earth.

Solution

2.5× 10 23 2.5× 10 23

Exercise 24

The farthest object astronomers have been able to see (as of 1981) is a quasar named 3C427. There seems to be a haze beyond this quasar that appears to mark the visual boundary of the universe. Quasar 3C427 is at a distance of 110,000,000,000,000,000,000,000,000 meters from the earth.

Exercise 25

The smallest known insects are about the size of a typical grain of sand. They are about 0.0002 0.0002 meters in length (2 ten-thousandths of a meter).

Solution

2× 10 4 2× 10 4

Exercise 26

Atoms such as hydrogen, carbon, nitrogen, and oxygen are about 0.0000000001 0.0000000001 meter across.

Exercise 27

The island of Manhattan, in New York, is about 57,000 square meters in area.

Solution

5.7× 10 4 5.7× 10 4

Exercise 28

The second largest moon of Saturn is Rhea. Rhea has a surface area of about 735,000 square meters, roughly the same surface area as Australia.

Exercise 29

A star, named Epsilon Aurigae B, has a diameter (distance across) of 2,800,000,000,000 meters. This diameter produces a surface area of about 24,630,000,000,000,000,000,000,000 square meters. This star is what astronomers call a red giant and it is the largest red giant known. If Epsilon Aurigae were placed at the sun’s position, its surface would extend out to the planet Uranus.

Solution

2.8× 10 12 ,2.463× 10 25 2.8× 10 12 ,2.463× 10 25

Exercise 30

The volume of the planet Venus is 927,590,000,000,000,000,000 cubic meters.

Exercise 31

The average mass of a newborn American female is about 3360 grams.

Solution

3.36× 10 3 3.36× 10 3

Exercise 32

The largest brain ever measured was that of a sperm whale. It had a mass of 9200 grams.

Exercise 33

The mass of the Eiffel tower in Paris, France, is 8,000,000 grams.

Solution

8× 10 6 8× 10 6

Exercise 34

In 1981, a Japanese company built the largest oil tanker to date. The ship has a mass of about 510,000,000,000 grams. This oil tanker is more than 6 times as massive as the U.S. aircraft carrier, U.S.S. Nimitz.

Exercise 35

In the constellation of Virgo, there is a cluster of about 2500 galaxies. The combined mass of these galaxies is 150,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 grams.

Solution

1.5× 10 62 1.5× 10 62

Exercise 36

The mass of an amoeba is about 0.000004 0.000004 gram.

Exercise 37

Cells in the human liver have masses of about 0.000000008 0.000000008 gram.

Solution

8× 10 9 8× 10 9

Exercise 38

The human sperm cell has a mass of about 0.000000000017 0.000000000017 gram.

Exercise 39

The principal protein of muscle is myosin. Myosin has a mass of 0.00000000000000000103 0.00000000000000000103 gram.

Solution

1.03× 10 18 1.03× 10 18

Exercise 40

Amino acids are molecules that combine to make up protein molecules. The amino acid tryptophan has a mass of 0.000000000000000000000340 0.000000000000000000000340 gram.

Exercise 41

An atom of the chemical element bromine has 35 electrons. The mass of a bromine atom is 0.000000000000000000000000031 0.000000000000000000000000031 gram.

Solution

3.1× 10 26 3.1× 10 26

Exercise 42

Physicists are performing experiments that they hope will determine the mass of a small particle called a neutrino. It is suspected that neutrinos have masses of about 0.0000000000000000000000000000001 0.0000000000000000000000000000001 gram.

Exercise 43

The approximate time it takes for a human being to die of asphyxiation is 316 seconds.

Solution

3.16× 10 2 3.16× 10 2

Exercise 44

On the average, the male housefly lives 1,468,800 seconds (17 days).

Exercise 45

Aluminum-26 has a half-life of 740,000 years.

Solution

7.4× 10 5 7.4× 10 5

Exercise 46

Manganese-53 has a half-life of 59,918,000,000,000 seconds (1,900,000 years).

Exercise 47

In its orbit around the sun, the earth moves a distance one and one half feet in about 0.0000316 0.0000316 second.

Solution

3.16× 10 5 3.16× 10 5

Exercise 48

A pi-meson is a subatomic particle that has a half-life of about 0.0000000261 0.0000000261 second.

Exercise 49

A subatomic particle called a neutral pion has a half-life of about 0.0000000000000001 0.0000000000000001 second.

Solution

1× 10 16 1× 10 16

Exercise 50

Near the surface of the earth, the speed of sound is 1195 feet per second.

For the following problems, convert the numbers from scientific notation to standard decimal form.

Exercise 51

The sun is about 1× 10 8 1× 10 8 meters from earth.

Solution

100,000,000 100,000,000

Exercise 52

The mass of the earth is about 5.98× 10 27 5.98× 10 27 grams.

Exercise 53

Light travels about 5.866× 10 12 5.866× 10 12 miles in one year.

Solution

5,866,000,000,000 5,866,000,000,000

Exercise 54

One year is about 3× 10 7 3× 10 7 seconds.

Exercise 55

Rubik’s cube has about 4.3× 10 19 4.3× 10 19 different configurations.

Solution

43,000,000,000,000,000,000 43,000,000,000,000,000,000

Exercise 56

A photon is a particle of light. A 100-watt light bulb emits 1× 10 20 1× 10 20 photons every second.

Exercise 57

There are about 6× 10 7 6× 10 7 cells in the retina of the human eye.

Solution

60,000,000 60,000,000

Exercise 58

A car traveling at an average speed will travel a distance about equal to the length of the smallest fingernail in 3.16× 10 4 3.16× 10 4 seconds.

Exercise 59

A ribosome of E. coli has a mass of about 4.7× 10 19 4.7× 10 19 grams.

Solution

0.00000000000000000047 0.00000000000000000047

Exercise 60

A mitochondrion is the energy-producing element of a cell. A mitochondrion is about 1.5× 10 6 1.5× 10 6 meters in diameter.

Exercise 61

There is a species of frogs in Cuba that attain a length of at most 1.25× 10 2 1.25× 10 2 meters.

Solution

0.0125 0.0125

Perform the following operations.

Exercise 62

(2× 10 4 )(3× 10 5 ) (2× 10 4 )(3× 10 5 )

Exercise 63

(4× 10 2 )(8× 10 6 ) (4× 10 2 )(8× 10 6 )

Solution

3.2× 10 9 3.2× 10 9

Exercise 64

(6× 10 14 )(6× 10 10 ) (6× 10 14 )(6× 10 10 )

Exercise 65

(3× 10 5 )(8× 10 7 ) (3× 10 5 )(8× 10 7 )

Solution

2.4× 10 3 2.4× 10 3

Exercise 66

(2× 10 1 )(3× 10 5 ) (2× 10 1 )(3× 10 5 )

Exercise 67

(9× 10 5 )(1× 10 11 ) (9× 10 5 )(1× 10 11 )

Solution

9× 10 16 9× 10 16

Exercise 68

(3.1× 10 4 )(3.1× 10 6 ) (3.1× 10 4 )(3.1× 10 6 )

Exercise 69

(4.2× 10 12 )(3.6× 10 20 ) (4.2× 10 12 )(3.6× 10 20 )

Solution

1.512× 10 31 1.512× 10 31

Exercise 70

(1.1× 10 6 ) 2 (1.1× 10 6 ) 2

Exercise 71

(5.9× 10 14 ) 2 (5.9× 10 14 ) 2

Solution

3.481× 10 29 3.481× 10 29

Exercise 72

(1.02× 10 17 ) 2 (1.02× 10 17 ) 2

Exercise 73

(8.8× 10 50 ) 2 (8.8× 10 50 ) 2

Solution

7.744× 10 99 7.744× 10 99

Exercise 74

If Mount Kilimanjaro was 1,000,000 times as high as it really is, how high would it be? (See problem 1.)

Exercise 75

If the planet Mars was 300,000 times as far from the sun as it really is, how far from the sun would it be? (See problem 2.)

Solution

6.687× 10 16 6.687× 10 16

Exercise 76

If 800,000,000 of the smallest insects known were lined up head to tail, how far would they stretch? (See problem 5.)

Exercise 77

If Rhea, the moon of Saturn, had a surface area 0.00000000002 0.00000000002 of its real surface area, what would that surface area be? (See problem 8.)

Solution

1.47× 10 5 1.47× 10 5

Exercise 78

If the star Epsilon Aurigae B had a surface area 0.005 0.005 of its real surface area, what would that surface area be? (See problem 9.)

Exercise 79

If the mass of all the galaxies in the constellation Virgo was only 0.0000000000000000000000003 0.0000000000000000000000003 of its real mass, what would that mass be? (See problem 15.)

Solution

4.5× 10 37 4.5× 10 37

Exercise 80

What is the mass of 15,000,000,000,000 bromine atoms? (See problem 21.)

Exercises for Review

Exercise 81

((Reference)) What integers can replace x x so that the statement 6<x<2 6<x<2 is true?

Solution

5,4,3 5,4,3

Exercise 82

((Reference)) Simplify (5 x 2 y 4 )(2x y 5 ) (5 x 2 y 4 )(2x y 5 )

Exercise 83

((Reference)) Determine the value of [(|5|)] [(|5|)] .

Solution

5 5

Exercise 84

((Reference)) Write x 3 y 5 z 4 x 3 y 5 z 4 so that only positive exponents appear.

Exercise 85

((Reference)) Write (2z+1) 3 (2z+1) 5 (2z+1) 3 (2z+1) 5 so that only positive exponents appear.

Solution

1 ( 2z+1 ) 2 1 ( 2z+1 ) 2

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