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Exponents

Module by: Community College Online Textbook Project. E-mail the author

Summary: Exponential notation is a short way of writing the same number multiplied by itself many times.

In this module, you will learn the short-cut to writing 2⋅2⋅2⋅22⋅2⋅2⋅2 size 12{2 cdot 2 cdot 2 cdot 2} {}. This is known as writing a number in exponential notation.

Definition of Exponential Notation

Exponential notation is a short way of writing the same number multiplied by itself many times.

Exponential notation uses a superscript for the number of times the number is repeated. The superscript is placed on the number to be multiplied (the factor), and is written like anan size 12{a rSup { size 8{n} } } {}where n is an integer and a can be any real number. a is called the base and n is called the exponent or power.

The nth power of a is defined as:

an=1⋅a⋅a⋅…⋅aan=1⋅a⋅a⋅…⋅a size 12{a rSup { size 8{n} } =1 cdot a cdot a cdot dotslow cdot a } {} (n times)

with a multiplied by itself n times.

The resulting value is called the argument.

For example, instead of 5⋅5⋅5⋅5⋅5⋅55⋅5⋅5⋅5⋅5⋅5 size 12{5 cdot 5 cdot 5 cdot 5 cdot 5 cdot 5} {}, we write 5656 size 12{5 rSup { size 8{6} } } {} to show that the number 5 is multiplied by itself 6 times.

5 is the base, and 6 is the exponent or power.

The result, 15625, is the argument.

56 is read as “five to the sixth power,” or more simply as “five to the sixth,” or “the sixth power of five.”

Likewise 5252 size 12{5 rSup { size 8{2} } } {} is 5⋅55⋅5 size 12{5 cdot 5} {} and 3535 size 12{3 rSup { size 8{5} } } {} is 3⋅3⋅3⋅3⋅33⋅3⋅3⋅3⋅3 size 12{3 cdot 3 cdot 3 cdot 3 cdot 3} {}. We will now have a closer look at writing numbers using exponential notation.

When a whole number is raised to the second power, it is said to be squared. The number 52 can be read as

  • 5 to the second power, or
  • 5 to the second, or
  • 5 squared.

When a whole number is raised to the third power, it is said to be cubed. The number 53 can be read as

  • 5 to the third power, or
  • 5 to the third, or
  • 5 cubed.

When a whole number is raised to the power of 4 or higher, we simply say that the number is raised to that particular power. The number 58 can be read as

  • 5 to the eighth power, or just
  • 5 to the eighth.

We can also define what it means if we have a negative index, -n. Then,

a−n=1÷a÷a÷…÷aa−n=1÷a÷a÷…÷a size 12{a rSup { size 8{ - n} } `=`1` div `a` div `a` div ` dotslow ` div `a } {}   (n times)

If n is an even integer, then anan size 12{a rSup { size 8{n} } } {} will always be positive for any non-zero real number a. For example, although -2 is negative, (−2)2=1⋅−2⋅−2=4(−2)2=1⋅−2⋅−2=4 size 12{ \( - 2 \) rSup { size 8{2} } =1 cdot - 2 cdot - 2=4} {}  is positive and so is (−2)−2=1÷−2÷−2=14(−2)−2=1÷−2÷−2=14 size 12{ \( - 2 \) rSup { size 8{ - 2} } =1 div ` - 2 div ` - 2= { { size 6{ size 10{1}} } over { size 10{4}} } } {}.

Examples, Exponential Notation

Write the following multiplication using exponents:

Example 1

3 · 3 

Since the factor 3 appears 2 times, we write this as

32

  

Example 2

62 · 62 · 62 · 62 · 62 · 62 · 62 · 62 · 62 

Since the factor 62 appears nine times, we write this as:

629

Expand each number (write without exponents):

Example 3

124.        The exponent 4 indicates that the base (12) is repeated 4 times, thus:

124 = 12 · 12 · 12 · 12

  

Example 4

7063.   The exponent 3 indicates that the base (706) is repeated 3 times in a multiplication.

7063 = 706 · 706 · 706

Exercises, Exponential Notation

Write each of the following using exponents:

Exercise 1

37 · 37

Solution

372

Exercise 2

16 · 16 · 16 · 16 · 16

Solution

165

Exercise 3

9 · 9 · 9 · 9 · 9 · 9 · 9 · 9 · 9 · 9

Solution

910

Write each of the following numbers without exponents:

Exercise 4

853

Solution

85 · 85 · 85

Exercise 5

47

Solution

4 · 4 · 4 · 4 · 4 · 4 · 4

Exercise 6

17392

Solution

1739 · 1739

Laws of Exponents

There are several laws we can use to make working with exponential numbers easier. We list all the laws here for easy reference.

a 0 = 1 a 0 = 1 size 12{a rSup { size 8{0} } =1} {}
(1)
a m × a n = a m + n a m × a n = a m + n size 12{a rSup { size 8{m} } times a rSup { size 8{n} } =a rSup { size 8{m+n} } } {}
(2)
a m ÷ a n = a m − n a m ÷ a n = a m − n size 12{a rSup { size 8{m} } ` div `a rSup { size 8{n} } `=`a rSup { size 8{m - n} } } {}
(3)
a − n = 1 a n a − n = 1 a n size 12{a rSup { size 8{ - n} } = { {1} over {a rSup { size 8{n} } } } } {}
(4)
ab n = a n b n ab n = a n b n size 12{ ital "ab" rSup { size 8{n} } =a rSup { size 8{n} } b rSup { size 8{n} } } {}
(5)
( a m ) n = a mn ( a m ) n = a mn size 12{ \( a rSup { size 8{m} } \) rSup { size 8{n} } =a rSup { size 8{ ital "mn"} } } {}
(6)

We explain each law in detail in the following sections.

Exponential Law 1

Our definition of exponential notation shows that:

a0=1a0=1 size 12{a rSup { size 8{0} } `=`1} {}, (a≠0)(a≠0) size 12{` \( a <> 0 \) } {}

For example, x0=1 and (1,000,000)0=1x0=1 and (1,000,000)0=1 size 12{x rSup { size 8{0} } `=``1" and " \( "1,000,000" \) rSup { size 8{0} } `=``1} {}.

Note that the base must be a non-zero value. 00 is called an indeterminate number, and has no value. This is because 00 = 0/0. If one considers 0 = 0 × n (where n can be any number) then it follows that 0/0 = n, where n can be any number – meaning the value of 0/0 cannot be determined.

Examples: Application using Exponential Law 1

  1. 16 0 = 1 16 0 = 1 size 12{"16" rSup { size 8{0} } =``1} {}
  2. 16 a 0 = 16 16 a 0 = 16 size 12{"16"a rSup { size 8{0} } =``"16"} {}
  3. ( 16 + a ) 0 = 1 ( 16 + a ) 0 = 1 size 12{ \( "16"+a \) rSup { size 8{0} } =``1} {}
  4. ( − 16 ) 0 = 1 ( − 16 ) 0 = 1 size 12{ \( - "16" \) rSup { size 8{0} } =``1} {}
  5. − 16 0 = − 1 − 16 0 = − 1 size 12{ - "16" rSup { size 8{0} } =`` - 1} {}

Exponential Law 2

Our definition of exponential notation shows that:

a m × a n = a m + n a m × a n = a m + n size 12{a rSup { size 8{m} } ` times `a rSup { size 8{n} } `=`a rSup { size 8{m+n} } } {}
(7)

That is:

am⋅an=1⋅a⋅…⋅aam⋅an=1⋅a⋅…⋅a size 12{a rSup { size 8{m} } cdot a rSup { size 8{n} } `=``1` cdot `a` cdot ` dotslow ` cdot `a } {} (m times) ⋅1⋅a⋅…⋅a ⋅1⋅a⋅…⋅a size 12{` cdot `1` cdot `a` cdot ` dotslow ` cdot ` ital "a "} {}  (n times)

             =1⋅a⋅…⋅a=1⋅a⋅…⋅a size 12{ {}= `1` cdot `a` cdot ` dotslow ` cdot `a" "``} {}    (m + n times)

             = am+n= am+n size 12{ {}= ital " a" rSup { size 8{m+n} } } {}

For example:

2 7 ⋅ 2 3 = ( 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ) ( 2 ⋅ 2 ⋅ 2 ) = 2 10 = 2 7 + 3 2 7 ⋅ 2 3 = ( 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ) ( 2 ⋅ 2 ⋅ 2 ) = 2 10 = 2 7 + 3 alignl { stack { size 12{`2 rSup { size 8{7} } cdot 2 rSup { size 8{3} } = \( 2 cdot 2 cdot 2 cdot 2 cdot 2 cdot 2 cdot 2 \) ital " " \( 2 cdot 2 cdot 2 \) } {} # `= 2 rSup { size 8{"10"} } {} # `= 2 rSup { size 8{7+3} } {} } } {}

This simple law illustrates the reason exponentials were originally invented. In the days before calculators, all multiplication had to be done by hand with a pencil and a pad of paper. Multiplication takes a very long time to do and is very tedious. Adding numbers, however, is easy and quick. This law says that adding the exponents of two exponential numbers (of the same base) is the same as multiplying the two numbers together. This means that, for certain numbers, there is no need to actually multiply the numbers together in order to find their multiple. This saved mathematicians a lot of time.

Examples: Application using Exponential Law 2

  1. x 2 ⋅ x 5 = x 7 x 2 ⋅ x 5 = x 7 size 12{x rSup { size 8{2} } cdot x rSup { size 8{5} } = ital " x" rSup { size 8{7} } } {}
  2. 2x 3 y ⋅ 5x 2 y 7 = 10 x 5 y 8 2x 3 y ⋅ 5x 2 y 7 = 10 x 5 y 8 size 12{2x rSup { size 8{3} } y cdot 5x rSup { size 8{2} } y rSup { size 8{7} } = "10"x rSup { size 8{5} } y rSup { size 8{8} } } {}
  3. 23⋅24=2723⋅24=27 size 12{2 rSup { size 8{3} } cdot 2 rSup { size 8{4} } = 2 rSup { size 8{7} } } {}    (Note that the base (2) stays the same.)
  4. 3 ⋅ 3 2a ⋅ 3 2 = 3 2a + 3 3 ⋅ 3 2a ⋅ 3 2 = 3 2a + 3 size 12{3 cdot 3 rSup { size 8{2a} } cdot 3 rSup { size 8{2} } =3 rSup { size 8{2a+3} } } {}

Exponential Law 3

a m ÷ a n = a m − n a m ÷ a n = a m − n size 12{a rSup { size 8{m} } `` div ``a rSup { size 8{n} } `=`a rSup { size 8{m - n} } } {}
(8)

We know from Law 2 that am+nam+n size 12{a rSup { size 8{m+n} } } {} is base a multiplied by itself m times plus a multiplied by itself n times. Law 3 extends this to the case where an exponent is negative.

a m a n = a ⋅ a ⋅ a ⋯ ⋅ a a ⋅ a ⋅ a ⋯ ⋅ a a m a n = a ⋅ a ⋅ a ⋯ ⋅ a a ⋅ a ⋅ a ⋯ ⋅ a size 12{ { {a rSup { size 8{m} } } over {a rSup { size 8{n} } } } `=` { {`a cdot a cdot a` dotsaxis ` cdot a`} over {a cdot a cdot a` dotsaxis ` cdot a} } } {} ( m times ) ( n times ) ( m times ) ( n times ) size 12{ { {` \( m`"times" \) `} over { \( n`"times" \) } } } {}

By factoring out anan size 12{a rSup { size 8{n} } } {} from both numerator and denominator, we are left with

     =a⋅a⋅a⋯⋅aa⋅a⋅a⋯⋅a=a⋅a⋅a⋯⋅aa⋅a⋅a⋯⋅a size 12{``=` { {`a cdot a cdot a dotsaxis cdot a`} over {`a cdot a cdot a dotsaxis cdot a`} } } {}(mtimes)(ntimes)(mtimes)(ntimes) size 12{ { {` \( m`"times" \) `} over { \( n`"times" \) } } } {}−a⋅a⋅a⋯⋅a−a⋅a⋅a⋯⋅a−a⋅a⋅a⋯⋅a−a⋅a⋅a⋯⋅a size 12{ { { - `a cdot a cdot a` dotsaxis cdot a} over { - `a cdot a cdot a` dotsaxis cdot a} } } {}(ntimes)(ntimes)(ntimes)(ntimes) size 12{ { {` \( n`"times" \) `} over { \( n`"times" \) } } } {}

     =a⋅a⋅a⋯⋅a=a⋅a⋅a⋯⋅a size 12{``=`a cdot a cdot a dotsaxis cdot a`} {}   (m – n times)

     =am−n=am−n size 12{``=`a rSup { size 8{m - n} } } {}

For example,

2 7 ÷ 2 3 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 2 ⋅ 2 ⋅ 2 = 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 4 = 2 7 − 3 2 7 ÷ 2 3 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 2 ⋅ 2 ⋅ 2 = 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 4 = 2 7 − 3 alignl { stack { size 12{`2 rSup { size 8{7} } div 2 rSup { size 8{3} } `=` { {2 cdot 2 cdot 2 cdot 2 cdot 2 cdot 2 cdot 2} over {2 cdot 2 cdot 2} } } {} # ```````````=``2 cdot 2 cdot 2 cdot 2 {} # ```````````=``2 rSup { size 8{4} } {} # ```````````=``2 rSup { size 8{7 - 3} } {} } } {}

Examples: Exponential Law 3

  1. a 6 a 2 = a 6 − 2 = a 4 a 6 a 2 = a 6 − 2 = a 4 size 12{ { {a rSup { size 8{6} } } over {a rSup { size 8{2} } } } `=`a rSup { size 8{6 - 2} } `=`a rSup { size 8{4} } } {}
  2. 3236=32−6=3−4=1343236=32−6=3−4=134 size 12{ { {3 rSup { size 8{2} } } over {3 rSup { size 8{6} } } } ``=``3 rSup { size 8{2 - 6} } ``=``3 rSup { size 8{ - 4} } `=` { {1} over {3 rSup { size 8{4} } } } ```} {}   (Always give the final answer with a positive index)
  3. 32 a 2 4a 8 = 8a − 6 = 8 a 6 32 a 2 4a 8 = 8a − 6 = 8 a 6 size 12{ { {"32"a rSup { size 8{2} } } over {4a rSup { size 8{8} } } } `=`8a rSup { size 8{ - 6} } `=` { {8} over {a rSup { size 8{6} } } } } {}
  4. a 3x a 4 = a 3x − 4 a 3x a 4 = a 3x − 4 size 12{ { {a rSup { size 8{3x} } } over {a rSup { size 8{4} } } } `=`a rSup { size 8{3x - 4} } } {}

Exponential Law 4

a − n = 1 a n , a ≠ 0 a − n = 1 a n , a ≠ 0 size 12{a rSup { size 8{ - n} } `= { {1} over {a rSup { size 8{n} } } } ,~`a <> 0} {}
(9)

Our definition of exponential notation for a negative exponent shows that

a−n=1÷a÷⋯÷aa−n=1÷a÷⋯÷a size 12{a rSup { size 8{ - n} } `=`1` div `a` div ` dotsaxis ` div `a} {}   (n times)

       =11⋅a⋅⋯⋅a=11⋅a⋅⋯⋅a size 12{ {}=` { {1} over {1` cdot `a` cdot ` dotsaxis ` cdot `a} } } {}(ntimes)(ntimes) size 12{ { {``} over { \( n`"times" \) } } } {} 

       =1an=1an size 12{ {}=` { {1} over {a rSup { size 8{n} } } } } {}

The minus sign in the exponent is just another way of writing that the whole exponential number is to be divided instead of multiplied.

For example, starting with Law 3, take the case of am−nam−n size 12{a rSup { size 8{m - n} } } {}, but where n > m:

2 2 − 9 = 2 2 2 9 = 2 ⋅ 2 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 1 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 1 2 7 = 2 − 7 2 2 − 9 = 2 2 2 9 = 2 ⋅ 2 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 1 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 1 2 7 = 2 − 7 alignl { stack { size 12{`2 rSup { size 8{2 - 9} } `=` { {2 rSup { size 8{2} } } over {2 rSup { size 8{9} } } } `} {} # ```````=` { {2` cdot `2} over {2` cdot `2` cdot `2` cdot `2` cdot `2` cdot `2` cdot `2` cdot `2` cdot `2} } {} # ```````= { {1} over {2 cdot 2 cdot 2 cdot 2 cdot 2 cdot 2 cdot 2} } {} # ```````= { {1} over {2 rSup { size 8{7} } } } {} # ```````=`2 rSup { size 8{ - 7} } {} } } {}

Examples: Exponential Law 4

  1. 2 − 2 = 1 2 2 = 1 4 2 − 2 = 1 2 2 = 1 4 size 12{2 rSup { size 8{ - 2} } = { {1} over {2 rSup { size 8{2} } } } = { {1} over {4} } } {}
  2. 2 − 2 3 2 = 1 2 2 ⋅ 3 2 = 1 36 2 − 2 3 2 = 1 2 2 ⋅ 3 2 = 1 36 size 12{ { {2 rSup { size 8{ - 2} } } over {3 rSup { size 8{2} } } } = { {1} over {2 rSup { size 8{2} } cdot 3 rSup { size 8{2} } } } = { {1} over {"36"} } } {}
  3. 2 3 − 3 = 3 2 3 = 27 8 2 3 − 3 = 3 2 3 = 27 8 size 12{ left ( { {2} over {3} } right ) rSup { size 8{ - 3} } = left ( { {3} over {2} } right ) rSup { size 8{3} } = { {"27"} over {8} } } {}
  4. m n − 4 = mn 4 m n − 4 = mn 4 size 12{ { {m} over {n rSup { size 8{ - 4} } } } = ital "mn" rSup { size 8{4} } } {}
  5. a − 3 ⋅ x 4 a 5 ⋅ x − 2 = x 4 ⋅ x 2 a 3 ⋅ a 5 = x 6 a 8 a − 3 ⋅ x 4 a 5 ⋅ x − 2 = x 4 ⋅ x 2 a 3 ⋅ a 5 = x 6 a 8 size 12{ { {a rSup { size 8{ - 3} } cdot x rSup { size 8{4} } } over {a rSup { size 8{5} } cdot x rSup { size 8{ - 2} } } } = { {x rSup { size 8{4} } cdot x rSup { size 8{2} } } over {a rSup { size 8{3} } cdot a rSup { size 8{5} } } } = { {x rSup { size 8{6} } } over {a rSup { size 8{8} } } } } {}

Exponential Law 5

( ab ) n = a n b n ( ab ) n = a n b n size 12{ \( ital "ab" \) rSup { size 8{n} } `=`a rSup { size 8{n} } b rSup { size 8{n} } } {}
(10)

The order in which two real numbers are multiplied together does not matter.

Therefore,

(ab)n=a⋅b⋅a⋅b⋅a⋅b⋅⋯⋅a⋅b(ab)n=a⋅b⋅a⋅b⋅a⋅b⋅⋯⋅a⋅b size 12{ \( ital "ab" \) rSup { size 8{n} } `=``a cdot b cdot a cdot b cdot a cdot b cdot `` dotsaxis ` cdot `a cdot b} {}    (n times)

         =a⋅a⋅…⋅a=a⋅a⋅…⋅a size 12{`=``a` cdot `a` cdot ` dotslow ` cdot `a} {} (n times) ⋅b⋅b⋅…⋅b⋅b⋅b⋅…⋅b size 12{` cdot `b` cdot `b` cdot ` dotslow ` cdot `b} {} (n times)

          =anbn=anbn size 12{ {}=``a rSup { size 8{n} } b rSup { size 8{n} } } {}

For example:

2 ⋅ 3 4 = ( 2 ⋅ 3 ) ⋅ ( 2 ⋅ 3 ) ⋅ ( 2 ⋅ 3 ) ⋅ ( 2 ⋅ 3 ) = ( 2 ⋅ 2 ⋅ 2 ⋅ 2 ) ⋅ ( 3 ⋅ 3 ⋅ 3 ⋅ 3 ) = 2 4 ⋅ 3 4 = 2 4 3 4 2 ⋅ 3 4 = ( 2 ⋅ 3 ) ⋅ ( 2 ⋅ 3 ) ⋅ ( 2 ⋅ 3 ) ⋅ ( 2 ⋅ 3 ) = ( 2 ⋅ 2 ⋅ 2 ⋅ 2 ) ⋅ ( 3 ⋅ 3 ⋅ 3 ⋅ 3 ) = 2 4 ⋅ 3 4 = 2 4 3 4 alignl { stack { size 12{`2` cdot 3 rSup { size 8{4} } = \( 2 cdot 3 \) cdot \( 2 cdot 3 \) cdot \( 2 cdot 3 \) cdot \( 2 cdot 3 \) } {} # `=`` \( 2 cdot 2 cdot 2 cdot 2 \) ` cdot ` \( 3 cdot 3 cdot 3 cdot 3 \) {} # `= 2 rSup { size 8{4} } ` cdot `3 rSup { size 8{4} } {} # `= 2 rSup { size 8{4} } 3 rSup { size 8{4} } {} } } {}
(11)

Examples: Exponential Law 5

  1. ( 2x 2 y ) 3 = 2 3 x 2 × 3 y 3 = 8x 6 y 3 ( 2x 2 y ) 3 = 2 3 x 2 × 3 y 3 = 8x 6 y 3 size 12{ \( 2x rSup { size 8{2} } y \) rSup { size 8{3} } `=`2 rSup { size 8{3} } x rSup { size 8{2 times 3} } y rSup { size 8{3} } `=`8x rSup { size 8{6} } y rSup { size 8{3} } } {}
  2. 7a b 3 2 = 49 a 2 b 6 7a b 3 2 = 49 a 2 b 6 size 12{ left ( { {7a} over {b rSup { size 8{3} } } } right )` rSup { size 8{2} } `=`` { {"49"a rSup { size 8{2} } } over {b rSup { size 8{6} } } } `} {}
  3. ( 5a n − 4 ) 3 = 125 a 3n − 12 ( 5a n − 4 ) 3 = 125 a 3n − 12 size 12{ \( 5a rSup { size 8{n - 4} } \) rSup { size 8{3} } `=`"125"a rSup { size 8{3n - "12"} } } {}

Exponential Law 6

( a m ) n = a mn ( a m ) n = a mn size 12{ \( a rSup { size 8{m} } \) rSup { size 8{n} } =a rSup { size 8{ ital "mn"} } } {}
(12)

We can find the exponential of an exponential just as well as we can for a number, because an exponential is a real number.

(am)n=am⋅am⋅am⋅…⋅am(am)n=am⋅am⋅am⋅…⋅am size 12{ \( a rSup { size 8{m} } \) rSup { size 8{n} } `=``a rSup { size 8{m} } ` cdot `a rSup { size 8{m} } ` cdot a rSup { size 8{m} } ` cdot `` dotslow ` cdot `a rSup { size 8{m} } } {}    (n times)

         =a⋅a⋅…⋅a =a⋅a⋅…⋅a size 12{`=``a cdot a cdot dotslow cdot ital "a " } {}      (m × n times)

          = amn= amn size 12{ {}= ital " a" rSup { size 8{ ital "mn"} } } {}

For example:

( 2 2 ) 3 = ( 2 2 ) ⋅ ( 2 2 ) ⋅ ( 2 2 ) = ( 2 ⋅ 2 ) ⋅ ( 2 ⋅ 2 ) ⋅ ( 2 ⋅ 2 ) = 2 6 = 2 2 × 3 ( 2 2 ) 3 = ( 2 2 ) ⋅ ( 2 2 ) ⋅ ( 2 2 ) = ( 2 ⋅ 2 ) ⋅ ( 2 ⋅ 2 ) ⋅ ( 2 ⋅ 2 ) = 2 6 = 2 2 × 3 alignl { stack { size 12{`` \( 2 rSup { size 8{2} } \) rSup { size 8{3} } = \( 2 rSup { size 8{2} } \) cdot \( 2 rSup { size 8{2} } \) cdot \( 2 rSup { size 8{2} } \) } {} # ``````````=`` \( 2 cdot 2 \) ` cdot ` \( 2 cdot 2 \) ` cdot ` \( 2 cdot 2 \) {} # ``````````= 2 rSup { size 8{6} } {} # ``````````= 2 rSup { size 8{2 times 3} } {} } } {}
(13)

Examples: Exponential Law 6

  1. ( x 3 ) 4 = x 12 ( x 3 ) 4 = x 12 size 12{ \( x rSup { size 8{3} } \) rSup { size 8{4} } `=`x rSup { size 8{"12"} } } {}
  2. [ ( a 4 ) 3 ] 2 = a 24 [ ( a 4 ) 3 ] 2 = a 24 size 12{ \[ \( a rSup { size 8{4} } \) rSup { size 8{3} } \] rSup { size 8{2} } `=``a rSup { size 8{"24"} } } {}
  3. ( 3 n + 3 ) 2 = 3 2n + 6 ( 3 n + 3 ) 2 = 3 2n + 6 size 12{ \( 3 rSup { size 8{n+3} } \) rSup { size 8{2} } `=`3 rSup { size 8{2n+6} } } {}

Module Review Exercises

Write the following examples using exponential notation.

Exercise 7

4 ⋅ 4 4 ⋅ 4 size 12{4` cdot `4} {}

Solution

42

Exercise 8

12 ⋅ 12 12 ⋅ 12 size 12{"12"` cdot `"12"} {}

Solution

122

Exercise 9

9 ⋅ 9 ⋅ 9 ⋅ 9 9 ⋅ 9 ⋅ 9 ⋅ 9 size 12{9` cdot `9` cdot `9` cdot `9} {}

Solution

94

Exercise 10

10 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 10 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 size 12{"10"` cdot `"10"` cdot `"10"` cdot `"10"` cdot `"10"` cdot `"10"} {}

Solution

106

Exercise 11

826 ⋅ 826 ⋅ 826 826 ⋅ 826 ⋅ 826 size 12{"826"` cdot `"826"` cdot `"826"} {}

Solution

8263

Exercise 12

3021 ⋅ 3021 ⋅ 3021 ⋅ 3021 3021 ⋅ 3021 ⋅ 3021 ⋅ 3021 size 12{"3021"` cdot `"3021"` cdot `"3021" cdot `"3021"} {}

Solution

30214

Exercise 13

6⋅6⋅6⋅⋯⋅66⋅6⋅6⋅⋯⋅6 size 12{6` cdot `6` cdot `6` cdot dotsaxis ` cdot `6} {}    (85 factors of 6).

Solution

685

Exercise 14

2⋅2⋅2⋅⋯⋅22⋅2⋅2⋅⋯⋅2 size 12{`2` cdot `2` cdot `2` cdot ` dotsaxis ` cdot `2} {}     (112 factors of 2).

Solution

2112

For the next examples, expand the terms. (Do not find the actual values).

Exercise 15

53

Solution

5 · 5 · 5

Exercise 16

152

Solution

15 · 15

Exercise 17

1175

Solution

117 · 117 · 117 · 117 · 117

Exercise 18

74

Solution

7 · 7 · 7 · 7

Determine the value of each of the powers.

Exercise 21

152

Solution

225

Exercise 22

34

Solution

81

Exercise 24

73

Solution

343

Simplify as far as possible.

Exercise 25

(2x)3

Solution

23 · x3 = 8x3

Exercise 26

3020

Solution

1

Exercise 27

(-2x)3

Solution

(-2)3 · x3 = -8x3

Exercise 28

23 · 24

Solution

23+4 = 27

Exercise 29

x 8 x 3 x 8 x 3 size 12{ { {x rSup { size 8{8} } } over {x rSup { size 8{3} } } } } {}

Solution

x 8 − 3 = x 5 x 8 − 3 = x 5 size 12{`x rSup { size 8{8 - 3} } `=`x rSup { size 8{5} } } {}

Exercise 30

25 x 2 5x 8 25 x 2 5x 8 size 12{` { {"25"x rSup { size 8{2} } } over {5x rSup { size 8{8} } } } } {}

Solution

25 5 x 2 − 8 = 5x − 6 = 5 x 6 25 5 x 2 − 8 = 5x − 6 = 5 x 6 size 12{` { {"25"} over {5} } x rSup { size 8{2 - 8} } `=`5x rSup { size 8{ - 6} } `=` { {5} over {x rSup { size 8{6} } } } } {}

Exercise 31

(3-1+2-1)-1

Solution

1 3 − 1 + 2 − 1 = 1 1 3 + 1 2 = 1 ⋅ 3 1 + 2 1 = 3 + 2 = 5 1 3 − 1 + 2 − 1 = 1 1 3 + 1 2 = 1 ⋅ 3 1 + 2 1 = 3 + 2 = 5 size 12{` { {1} over {3 rSup { size 8{ - 1} } `+`2 rSup { size 8{ - 1} } } } `=` { {1} over { { {1} over {3} } `+` { {1} over {2} } } } `=`1` cdot ` left ( { {3} over {1} } `+` { {2} over {1} } right )`=`3`+`2`=`5} {}

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