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1.5 Rules of Exponents and Scientific Notation

Learning Objectives

  1. Review the rules of exponents.
  2. Review the definition of negative exponents and zero as an exponent.
  3. Work with numbers using scientific notation.

Review of the Rules of Exponents

In this section, we review the rules of exponents. Recall that if a factor is repeated multiple times, then the product can be written in exponential form xn. The positive integer exponent n indicates the number of times the base x is repeated as a factor.

Consider the product of x4 and x6,

Expanding the expression using the definition produces multiple factors of the base which is quite cumbersome, particularly when n is large. For this reason, we have useful rules to help us simplify expressions with exponents. In this example, notice that we could obtain the same result by adding the exponents.

x4x6=x4+6=x10Productruleforexponents

In general, this describes the product rule for exponentsxmxn=xm+n; the product of two expressions with the same base can be simplified by adding the exponents.. In other words, when multiplying two expressions with the same base we add the exponents. Compare this to raising a factor involving an exponent to a power, such as (x6)4.

Here we have 4 factors of x6, which is equivalent to multiplying the exponents.

(x6)4=x64=x24Powerruleforexponents

This describes the power rule for exponents(xm)n=xmn; a power raised to a power can be simplified by multiplying the exponents.. Now we consider raising grouped products to a power. For example,

(x2y3)4=x2y3x2y3x2y3x2y3=x2x2x2x2y3y3y3y3      Commutativeproperty=x2+2+2+2y3+3+3+3=x8y12

After expanding, we are left with four factors of the product x2y3. This is equivalent to raising each of the original grouped factors to the fourth power and applying the power rule.

(x2y3)4=(x2)4(y3)4=x8y12

In general, this describes the use of the power rule for a product as well as the power rule for exponents. In summary, the rules of exponents streamline the process of working with algebraic expressions and will be used extensively as we move through our study of algebra. Given any positive integers m and n where x,y0 we have

Product rule for exponents:

xmxn=xm+n

Quotient rule for exponents:

xmxn=xmn

Power rule for exponents:

(xm)n=xmn

Power rule for a product:

(xy)n=xnyn

Power rule for a quotient:

(xy)n=xnyn

(xy)n=xnyn; if a product is raised to a power, then apply that power to each factor in the product.

(xy)n=xnyn; if a quotient is raised to a power, then apply that power to the numerator and the denominator.

These rules allow us to efficiently perform operations with exponents.

Example 1

Simplify: 1041012103.

Solution:

1041012103=1016103          Productrule=10163        Quotientrule=1013

Answer: 1013

In the previous example, notice that we did not multiply the base 10 times itself. When applying the product rule, add the exponents and leave the base unchanged.

Example 2

Simplify: (x5x4x)2.

Solution:

Recall that the variable x is assumed to have an exponent of one, x=x1.

(x5x4x)2=(x5+4+1)2=(x10)2=x102=x20

Answer: x20

The base could in fact be any algebraic expression.

Example 3

Simplify: (x+y)9(x+y)13.

Solution:

Treat the expression (x+y) as the base.

(x+y)9(x+y)13=(x+y)9+13=(x+y)22

Answer: (x+y)22

The commutative property of multiplication allows us to use the product rule for exponents to simplify factors of an algebraic expression.

Example 4

Simplify: 8x5y3x7y3.

Solution:

Multiply the coefficients and add the exponents of variable factors with the same base.

8x5y3x7y3=83x5x7y1y3      Commutativeproperty=24x5+7y1+3              Powerruleforexponents=24x12y4

Answer: 24x12y4

Division involves the quotient rule for exponents.

Example 5

Simplify: 33x7y5(xy)1011x6y(xy)3.

Solution:

33x7y5(xy)1011x6y(xy)3=3311x76y51(xy)103=3x1y4(xy)7

Answer: 3xy4(xy)7

The power rule for a quotient allows us to apply that exponent to the numerator and denominator. This rule requires that the denominator is nonzero and so we will make this assumption for the remainder of the section.

Example 6

Simplify: (4a2bc4)3.

Solution:

First apply the power rule for a quotient and then the power rule for a product.

(4a2bc4)3=(4a2b)3(c4)3            Powerruleforaquotient=(4)3(a2)3(b)3(c4)3           Powerruleforaproduct=64a6b3c12

Answer: 64a6b3c12

Using the quotient rule for exponents, we can define what it means to have zero as an exponent. Consider the following calculation:

1=2525=5252=522=50

Twenty-five divided by twenty-five is clearly equal to one, and when the quotient rule for exponents is applied, we see that a zero exponent results. In general, given any nonzero real number x and integer n,

1=xnxn=xnn=x0

This leads us to the definition of zero as an exponentx0=1; any nonzero base raised to the 0 power is defined to be 1.,

x0=1x0

It is important to note that 00 is indeterminate. If the base is negative, then the result is still positive one. In other words, any nonzero base raised to the zero power is defined to be equal to one. In the following examples assume all variables are nonzero.

Example 7

Simplify:

  1. (2x)0
  2. 2x0

Solution:

  1. Any nonzero quantity raised to the zero power is equal to 1.

    (2x)0=1

  2. In the example, 2x0, the base is x, not −2x.

    2x0=2x0=21=2

Noting that 20=1 we can write,

123=2023=203=23

In general, given any nonzero real number x and integer n,

1xn=x0xn=x0n=xnx0

This leads us to the definition of negative exponentsxn=1xn, given any integer n, where x is nonzero.:

xn=1xnx0

An expression is completely simplified if it does not contain any negative exponents.

Example 8

Simplify: (4x2y)2.

Solution:

Rewrite the entire quantity in the denominator with an exponent of 2 and then simplify further.

(4x2y)2=1(4x2y)2=1(4)2(x2)2(y)2=116x4y2

Answer: 116x4y2

Sometimes negative exponents appear in the denominator.

Example 9

Simplify: x3y4.

Solution:

x3y4=1x31y4=1x3y41=y4x3

Answer: y4x3

The previous example suggests a property of quotients with negative exponentsxnym=ymxn, given any integers m and n, where x0 and y0.. Given any integers m and n where x0 and y0, then

xnym=1xn1ym=1xnym1=ymxn

This leads us to the property

xnym=ymxn

In other words, negative exponents in the numerator can be written as positive exponents in the denominator and negative exponents in the denominator can be written as positive exponents in the numerator.

Example 10

Simplify: 5x3y3z4.

Solution:

Take care with the coefficient −5, recognize that this is the base and that the exponent is actually positive one: 5=(5)1. Hence, the rules of negative exponents do not apply to this coefficient; leave it in the numerator.

5x3y3z4=5x3y3z4=5y3z4x3

Answer: 5y3z4x3

In summary, given integers m and n where x,y0 we have

Zero exponent:

x0=1

Negative exponent:

xn=1xn

Quotients with negative exponents:

xnym=ymxn

Furthermore, all of the rules of exponents defined so far extend to any integer exponents. We will expand the scope of these properties to include any real number exponents later in the course.

Try this! Simplify: (2x2y3z)4.

Answer: x8z416y12

Scientific Notation

Real numbers expressed using scientific notationReal numbers expressed the form a×10n, where n is an integer and 1a<10. have the form, a×10n where n is an integer and 1a<10. This form is particularly useful when the numbers are very large or very small. For example,

9,460,000,000,000,000m=9.46×1015m      Onelightyear0.000000000025m=2.5×1011mRadiusofahydrogenatom

It is cumbersome to write all the zeros in both of these cases. Scientific notation is an alternative, compact representation of these numbers. The factor 10n indicates the power of ten to multiply the coefficient by to convert back to decimal form:

This is equivalent to moving the decimal in the coefficient fifteen places to the right.

A negative exponent indicates that the number is very small:

This is equivalent to moving the decimal in the coefficient eleven places to the left.

 

Converting a decimal number to scientific notation involves moving the decimal as well. Consider all of the equivalent forms of 0.00563 with factors of 10 that follow:

0.00563=0.0563×101=0.563×102=5.63×103=56.3×104=563×105

While all of these are equal, 5.63×103 is the only form expressed in correct scientific notation. This is because the coefficient 5.63 is between 1 and 10 as required by the definition. Notice that we can convert 5.63×103 back to decimal form, as a check, by moving the decimal three places to the left.

xmxn=xmn; the quotient of two expressions with the same base can be simplified by subtracting the exponents.

Example 11

Write 1,075,000,000,000 using scientific notation.

Solution:

Here we count twelve decimal places to the left of the decimal point to obtain the number 1.075.

1,075,000,000,000=1.075×1012

Answer: 1.075×1012

Example 12

Write 0.000003045 using scientific notation.

Solution:

Here we count six decimal places to the right to obtain 3.045.

0.000003045=3.045×106

Answer: 3.045×106

Often we will need to perform operations when using numbers in scientific notation. All the rules of exponents developed so far also apply to numbers in scientific notation.

Example 13

Multiply: (4.36×105)(5.3×1012).

Solution:

Use the fact that multiplication is commutative, and apply the product rule for exponents.

(4.36×105)(5.30×1012)=(4.365.30)×(1051012)=23.108×105+12=2.3108×101×107=2.3108×101+7=2.3108×108

Answer: 2.3108×108

Example 14

Divide: (3.24×108)÷(9.0×103).

Solution:

(3.24×108)(9.0×103)=(3.249.0)×(108103)=0.36×108(3)=0.36×108+3=3.6×101×1011=3.6×101+11=3.6×1010

Answer: 3.6×1010

Example 15

The speed of light is approximately 6.7×108 miles per hour. Express this speed in miles per second.

Solution:

A unit analysis indicates that we must divide the number by 3,600.

6.7×108milesperhour=6.7×108miles1hour(1hour60minutes)(1minutes60seconds)=6.7×108miles3600seconds=(6.73600)×1080.0019×108         roundedtotwosignificantdigits=1.9×103×108=1.9×103+8=1.9×105

Answer: The speed of light is approximately 1.9×105 miles per second.

Example 16

The Sun moves around the center of the galaxy in a nearly circular orbit. The distance from the center of our galaxy to the Sun is approximately 26,000 light-years. What is the circumference of the orbit of the Sun around the galaxy in meters?

Solution:

One light-year measures 9.46×1015 meters. Therefore, multiply this by 26,000 or 2.60×104 to find the length of 26,000 light years in meters.

(9.46×1015)(2.60×104)=9.462.60×101510424.6×1019=2.46×1011019=2.46×1020

The radius r of this very large circle is approximately 2.46×1020 meters. Use the formula C=2πr to calculate the circumference of the orbit.

C=2πr2(3.14)(2.46×1020)=15.4×1020=1.54×1011020=1.54×1021

Answer: The circumference of the Sun’s orbit is approximately 1.54×1021 meters.

Try this! Divide: (3.15×105)÷(12×1013).

Answer: 2.625×107

Key Takeaways

  • When multiplying two quantities with the same base, add exponents: xmxn=xm+n.
  • When dividing two quantities with the same base, subtract exponents: xmxn=xmn.
  • When raising powers to powers, multiply exponents: (xm)n=xmn.
  • When a grouped quantity involving multiplication and division is raised to a power, apply that power to all of the factors in the numerator and the denominator: (xy)n=xnyn and (xy)n=xnyn.
  • Any nonzero quantity raised to the 0 power is defined to be equal to 1: x0=1.
  • Expressions with negative exponents in the numerator can be rewritten as expressions with positive exponents in the denominator: xn=1xn.
  • Expressions with negative exponents in the denominator can be rewritten as expressions with positive exponents in the numerator: 1xm=xm.
  • Take care to distinguish negative coefficients from negative exponents.
  • Scientific notation is particularly useful when working with numbers that are very large or very small.

Topic Exercises

    Part A: Rules of Exponents

      Simplify. (Assume all variables represent nonzero numbers.)

    1. 104107

    2. 7372

    3. 102104105
    4. 757972
    5. x3x2

    6. y5y3

    7. a8a6a5
    8. b4b10b8
    9. x2nx3nxn
    10. xnx8nx3n
    11. (x5)3

    12. (y4)3

    13. (x4y5)3

    14. (x7y)5

    15. (x2y3z4)4

    16. (xy2z3)2

    17. (5x2yz3)2

    18. (2xy3z4)5

    19. (x2yz5)n

    20. (xy2z3)2n

    21. (xx3x2)3

    22. (y2y5y)2

    23. a2(a4)2a3
    24. aa3a2(a2)3
    25. (2x+3)4(2x+3)9

    26. (3y1)7(3y1)2

    27. (a+b)3(a+b)5

    28. (x2y)7(x2y)3

    29. 5x2y3xy2

    30. 10x3y22xy

    31. 6x2yz33xyz4

    32. 2xyz2(4x2y2z)

    33. 3xny2n5x2y

    34. 8x5nyn2x2ny

    35. 40x5y3z4x2y2z
    36. 8x2y5z316x2yz
    37. 24a8b3(a5b)108a5b3(a5b)2
    38. 175m9n5(m+n)725m8n(m+n)3
    39. (2x4y2z)6
    40. (3xy4z7)5
    41. (3ab22c3)3
    42. (10a3b3c2)2
    43. (2xy4z3)4
    44. (7x9yz4)3
    45. (xy2z3)n
    46. (2x2y3z)n
    47. (5x)0

    48. (3x2y)0

    49. 5x0

    50. 3x2y0

    51. (2a2b0c3)5

    52. (3a4b2c0)4

    53. (9x3y2z0)23xy2
    54. (5x0y5z)325y2z0
    55. 2x3

    56. (2x)2

    57. a4a5a2

    58. b8b3b4

    59. a8a3a6
    60. b10b4b2
    61. 10x3y2

    62. 3x5y2

    63. 3x2y2z1

    64. 5x4y2z2

    65. 25x3y25x1y3
    66. 9x1y3z53x2y2z1
    67. (5x3y2z)3
    68. (7x2y5z2)2
    69. (2x3zy2)5
    70. (5x5z22y3)3
    71. (12x3y2z2x7yz8)3
    72. (150xy8z290x7y2z)2
    73. (9a3b4c23a3b5c7)4
    74. (15a7b5c83a6b2c3)3

      The value in dollars of a new mobile phone can be estimated by using the formula V=210(2t+1)1, where t is the number of years after purchase.

    1. How much was the phone worth new?

    2. How much will the phone be worth in 1 year?

    3. How much will the phone be worth in 3 years?

    4. How much will the phone be worth in 10 years?

    5. How much will the phone be worth in 100 years?

    6. According to the formula, will the phone ever be worthless? Explain.

    7. The height of a particular right circular cone is equal to the square of the radius of the base, h=r2. Find a formula for the volume in terms of r.

    8. A sphere has a radius r=3x2. Find the volume in terms of x.

    Part B: Scientific Notation

      Convert to a decimal number.

    1. 5.2×108

    2. 6.02×109

    3. 1.02×106

    4. 7.44×105

      Rewrite using scientific notation.

    1. 7,050,000

    2. 430,000,000,000

    3. 0.00005001

    4. 0.000000231

      Perform the operations.

    1. (1.2×109)(3×105)

    2. (4.8×105)(1.6×1020)

    3. (9.1×1023)(3×1010)

    4. (5.5×1012)(7×1025)

    5. 9.6×10161.2×104
    6. 4.8×10142.4×106
    7. 4×1088×1010
    8. 2.3×10239.2×103
    9. 987,000,000,000,000×23,000,000

    10. 0.00000000024×0.00000004

    11. 0.000000000522÷0.0000009

    12. 81,000,000,000÷0.0000648

    13. The population density of Earth refers to the number of people per square mile of land area. If the total land area on Earth is 5.751×107 square miles and the population in 2007 was estimated to be 6.67×109 people, then calculate the population density of Earth at that time.

    14. In 2008 the population of New York City was estimated to be 8.364 million people. The total land area is 305 square miles. Calculate the population density of New York City.

    15. The mass of Earth is 5.97×1024 kilograms and the mass of the Moon is 7.35×1022 kilograms. By what factor is the mass of Earth greater than the mass of the Moon?

    16. The mass of the Sun is 1.99×1030 kilograms and the mass of Earth is 5.97×1024 kilograms. By what factor is the mass of the Sun greater than the mass of Earth? Express your answer in scientific notation.

    17. The radius of the Sun is 4.322×105 miles and the average distance from Earth to the Moon is 2.392×105 miles. By what factor is the radius of the Sun larger than the average distance from Earth to the Moon?

    18. One light year, 9.461×1015 meters, is the distance that light travels in a vacuum in one year. If the distance from our Sun to the nearest star, Proxima Centauri, is estimated to be 3.991×1016 meters, then calculate the number of years it would take light to travel that distance.

    19. It is estimated that there are about 1 million ants per person on the planet. If the world population was estimated to be 6.67 billion people in 2007, then estimate the world ant population at that time.

    20. The radius of the earth is 6.3×106 meters and the radius of the sun is 7.0×108 meters. By what factor is the radius of the Sun larger than the radius of the Earth?

    21. A gigabyte is 1×109 bytes and a megabyte is 1×106 bytes. If the average song in the MP3 format consumes about 4.5 megabytes of storage, then how many songs will fit on a 4-gigabyte memory card?

    22. Water weighs approximately 18 grams per mole. If one mole is about 6×1023 molecules, then approximate the weight of each molecule of water.

    Part C: Discussion Board

    1. Use numbers to show that (x+y)nxn+yn.

    2. Why is 00 indeterminate?

    3. Explain to a beginning algebra student why 222345.

    4. René Descartes (1637) established the usage of exponential form: a2, a3, and so on. Before this, how were exponents denoted?

Answers

  1. 1011

  2. 10

  3. x5

  4. a9

  5. x4n

  6. x15

  7. x12y15

  8. x8y12z16

  9. 25x4y2z6

  10. x2nynz5n

  11. x18

  12. a7

  13. (2x+3)13

  14. (a+b)8

  15. 15x3y3

  16. 18x3y2z7

  17. 15xn+2y2n+1

  18. 10x3y

  19. 3a3(a5b)8

  20. 64x24y12z6

  21. 27a3b68c9
  22. 16x4y16z12
  23. xny2nz3n
  24. 1

  25. −5

  26. 32a10c15

  27. 27x5y2

  28. 2x3
  29. a

  30. a11

  31. 10y2x3
  32. 3y2x2z
  33. 5y5x2
  34. x9125y6z3
  35. x15y1032z5
  36. 216y3x12z21
  37. a24b481c20
  38. $210

  39. $30

  40. $1.04

  41. V=13πr4

  1. 520,000,000

  2. 0.00000102

  3. 7.05×106

  4. 5.001×105

  5. 3.6×1014

  6. 2.73×1034

  7. 8×1020

  8. 5×1019

  9. 2.2701×1022

  10. 5.8×104

  11. About 116 people per square mile

  12. 81.2

  13. 1.807

  14. 6.67×1015 ants

  15. Approximately 889 songs

  1. Answer may vary

  2. Answer may vary