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Introduction

In this chapter, you will learn about the short cuts to writing 2×2×2×22×2×2×2 . This is known as writing a number in exponential notation.

Definition

Exponential notation is a short way of writing the same number multiplied by itself many times. For example, instead of 5×5×55×5×5, we write 5353 to show that the number 5 is multiplied by itself 3 times and we say “5 to the power of 3”. Likewise 5252 is 5×55×5 and 3535 is 3×3×3×3×33×3×3×3×3. We will now have a closer look at writing numbers using exponential notation.

Definition 1: Exponential Notation

Exponential notation means a number written like

a n a n
(1)

where nn is an integer and aa can be any real number. aa is called the base and nn is called the exponent or index.

The nthnth power of aa is defined as:

a n = a × a × ⋯ × a ( n times ) a n = a × a × ⋯ × a ( n times )
(2)

with aa multiplied by itself nn times.

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

a - n = 1 a × a × ⋯ × a ( n times ) a - n = 1 a × a × ⋯ × a ( n times )
(3)

Tip:

Exponentials

If nn is an even integer, then anan will always be positive for any non-zero real number aa. For example, although -2-2 is negative, (-2)2=-2×-2=4(-2)2=-2×-2=4 is positive and so is (-2)-2=1-2×-2=14(-2)-2=1-2×-2=14.

Figure 1
Khan Academy video on Exponents - 1

Figure 2
Khan Academy video on Exponents-2

Laws of Exponents

There are several laws we can use to make working with exponential numbers easier. Some of these laws might have been seen in earlier grades, but we will list all the laws here for easy reference and explain each law in detail, so that you can understand them and not only remember them.

a 0 = 1 a m × a n = a m + n a - n = 1 a n a m ÷ a n = a m - n ( a b ) n = a n b n ( a m ) n = a m n a 0 = 1 a m × a n = a m + n a - n = 1 a n a m ÷ a n = a m - n ( a b ) n = a n b n ( a m ) n = a m n
(4)

Exponential Law 1: a0=1a0=1

Our definition of exponential notation shows that

a 0 = 1 , ( a ≠ 0 ) a 0 = 1 , ( a ≠ 0 )
(5)

To convince yourself of why this is true, use the fourth exponential law above (division of exponents) and consider what happens when m=nm=n.

For example, x0=1x0=1 and (1000000)0=1(1000000)0=1.

Application using Exponential Law 1: a0=1,(a≠0)a0=1,(a≠0)

  1. 16 0 16 0
  2. 16 a 0 16 a 0
  3. ( 16 + a ) 0 ( 16 + a ) 0
  4. ( - 16 ) 0 ( - 16 ) 0
  5. - 16 0 - 16 0

    Click here for the solution

Exponential Law 2: am×an=am+nam×an=am+n

Figure 3
Khan Academy video on Exponents - 3

Our definition of exponential notation shows that

a m × a n = 1 × a × ... × a ( m times ) × 1 × a × ... × a ( n times ) = 1 × a × ... × a ( m + n times ) = a m + n a m × a n = 1 × a × ... × a ( m times ) × 1 × a × ... × a ( n times ) = 1 × a × ... × a ( m + n times ) = a m + n
(6)

For example,

2 7 × 2 3 = ( 2 × 2 × 2 × 2 × 2 × 2 × 2 ) × ( 2 × 2 × 2 ) = 2 7 + 3 = 2 10 2 7 × 2 3 = ( 2 × 2 × 2 × 2 × 2 × 2 × 2 ) × ( 2 × 2 × 2 ) = 2 7 + 3 = 2 10
(7)

Note: Interesting Fact :

This simple law is the reason why 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 very easy and quick to do. If you look at what this law is saying you will realise that it means that adding the exponents of two exponential numbers (of the same base) is the same as multiplying the two numbers together. This meant that for certain numbers, there was no need to actually multiply the numbers together in order to find out what their multiple was. This saved mathematicians a lot of time, which they could use to do something more productive.

Application using Exponential Law 2: am×an=am+nam×an=am+n

  1. x 2 · x 5 x 2 · x 5
  2. 2 3 · 2 4 2 3 · 2 4 [Take note that the base (2) stays the same.]
  3. 3 × 3 2 a × 3 2 3 × 3 2 a × 3 2

    Click here for the solution

Exponential Law 3: a-n=1an,a≠0a-n=1an,a≠0

Our definition of exponential notation for a negative exponent shows that

a - n = 1 ÷ a ÷ ... ÷ a ( n times ) = 1 1 × a × ⋯ × a ( n times ) = 1 a n a - n = 1 ÷ a ÷ ... ÷ a ( n times ) = 1 1 × a × ⋯ × a ( n times ) = 1 a n
(8)

This means that a minus sign in the exponent is just another way of showing that the whole exponential number is to be divided instead of multiplied.

For example,

2 - 7 = 1 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1 2 7 2 - 7 = 1 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1 2 7
(9)

This law is useful in helping us simplify fractions where there are exponents in both the denominator and the numerator. For example:

a-3 a4 = 1a3a4 = 1a7 a-3 a4 = 1a3a4 = 1a7
(10)

Application using Exponential Law 3: a-n=1an,a≠0a-n=1an,a≠0

  1. 2 - 2 = 1 2 2 2 - 2 = 1 2 2
  2. 2 - 2 3 2 2 - 2 3 2
  3. ( 2 3 ) - 3 ( 2 3 ) - 3
  4. m n - 4 m n - 4
  5. a - 3 · x 4 a 5 · x - 2 a - 3 · x 4 a 5 · x - 2

    Click here for the solution

Exponential Law 4: am÷an=am-nam÷an=am-n

We already realised with law 3 that a minus sign is another way of saying that the exponential number is to be divided instead of multiplied. Law 4 is just a more general way of saying the same thing. We get this law by multiplying law 3 by amam on both sides and using law 2.

a m a n = a m a - n = a m - n a m a n = a m a - n = a m - n
(11)

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
(12)

Figure 4
Khan academy video on exponents - 4

Application using Exponential Law 4: am÷an=am-nam÷an=am-n

  1. a 6 a 2 = a 6 - 2 a 6 a 2 = a 6 - 2
  2. 3 2 3 6 3 2 3 6
  3. 32 a 2 4 a 8 32 a 2 4 a 8
  4. a 3 x a 4 a 3 x a 4

    Click here for the solution

Exponential Law 5: (ab)n=anbn(ab)n=anbn

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

( a b ) n = a × b × a × b × ... × a × b ( n times ) = a × a × ... × a ( n times ) × b × b × ... × b ( n times ) = a n b n ( a b ) n = a × b × a × b × ... × a × b ( n times ) = a × a × ... × a ( n times ) × b × b × ... × b ( n times ) = a n b n
(13)

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
(14)

Application using Exponential Law 5: (ab)n=anbn(ab)n=anbn

  1. ( 2 x y ) 3 = 2 3 x 3 y 3 ( 2 x y ) 3 = 2 3 x 3 y 3
  2. ( 7 a b ) 2 ( 7 a b ) 2
  3. ( 5 a ) 3 ( 5 a ) 3

    Click here for the solution

Exponential Law 6: (am)n=amn(am)n=amn

We can find the exponential of an exponential of a number. An exponential of a number is just a real number. So, even though the sentence sounds complicated, it is just saying that you can find the exponential of a number and then take the exponential of that number. You just take the exponential twice, using the answer of the first exponential as the argument for the second one.

( a m ) n = a m × a m × ... × a m ( n times ) = a × a × ... × a ( m × n times ) = a m n ( a m ) n = a m × a m × ... × a m ( n times ) = a × a × ... × a ( m × n times ) = a m n
(15)

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 )
(16)

Application using Exponential Law 6: (am)n=amn(am)n=amn

  1. ( x 3 ) 4 ( x 3 ) 4
  2. [ ( a 4 ) 3 ] 2 [ ( a 4 ) 3 ] 2
  3. ( 3 n + 3 ) 2 ( 3 n + 3 ) 2

    Click here for the solution

Exercise 1: Simplifying indices

Simplify: 52x-1·9x-2152x-352x-1·9x-2152x-3

Solution
  1. Step 1. Factorise all bases into prime factors: :
    = 5 2 x - 1 · ( 3 2 ) x - 2 ( 5 . 3 ) 2 x - 3 = 5 2 x - 1 · 3 2 x - 4 5 2 x - 3 · 3 2 x - 3 = 5 2 x - 1 · ( 3 2 ) x - 2 ( 5 . 3 ) 2 x - 3 = 5 2 x - 1 · 3 2 x - 4 5 2 x - 3 · 3 2 x - 3
    (17)
  2. Step 2. Add and subtract the indices of the same bases as per laws 2 and 4: :
    = 5 2 x - 1 - 2 x + 3 · 3 2 x - 4 - 2 x + 3 = 5 2 · 3 - 1 = 5 2 x - 1 - 2 x + 3 · 3 2 x - 4 - 2 x + 3 = 5 2 · 3 - 1
    (18)
  3. Step 3. Write the simplified answer with positive indices: :
    = 25 3 = 25 3
    (19)

Investigation : Exponential Numbers

Match the answers to the questions, by filling in the correct answer into the Answer column. Possible answers are: 3232, 1, -1-1, -13-13, 8. Answers may be repeated.

Table 1
Question Answer
2 3 2 3  
7 3 - 3 7 3 - 3  
( 2 3 ) - 1 ( 2 3 ) - 1  
8 7 - 6 8 7 - 6  
( - 3 ) - 1 ( - 3 ) - 1  
( - 1 ) 23 ( - 1 ) 23  

We will use all these laws in Equations and Inequalities to help us solve exponential equations.

The following video gives an example on using some of the concepts covered in this chapter.

Figure 5
Khan Academy video on Exponents - 5

Summary

  • Exponential notation means a number written like a n a n where nn is an integer and aa can be any real number.
  • aa is called the base and nn is called the exponent or index.
  • The nthnth power of aa is defined as: a n = a × a × ⋯ × a ( n times ) a n = a × a × ⋯ × a ( n times )
  • There are six laws of exponents:
    • Exponential Law 1: a0=1a0=1
    • Exponential Law 2: am×an=am+nam×an=am+n
    • Exponential Law 3: a-n=1an,a≠0a-n=1an,a≠0
    • Exponential Law 4: am÷an=am-nam÷an=am-n
    • Exponential Law 5: (ab)n=anbn(ab)n=anbn
    • Exponential Law 6: (am)n=amn(am)n=amn

End of Chapter Exercises

  1. Simplify as far as possible:
    1. 30203020
    2. 1010
    3. (xyz)0(xyz)0
    4. [(3x4y7z12)5(-5x9y3z4)2]0[(3x4y7z12)5(-5x9y3z4)2]0
    5. (2x)3(2x)3
    6. (-2x)3(-2x)3
    7. (2x)4(2x)4
    8. (-2x) 4(-2x) 4
    Click here for the solution
  2. Simplify without using a calculator. Leave your answers with positive exponents.
    1. 3x-3(3x)23x-3(3x)2
    2. 5x0+8-2-(12)-2·1x5x0+8-2-(12)-2·1x
    3. 5b-35b+15b-35b+1
    Click here for the solution
  3. Simplify, showing all steps:
    1. 2a-2.3a+36a2a-2.3a+36a
    2. a2m+n+pam+n+p·ama2m+n+pam+n+p·am
    3. 3n·9n-327n-13n·9n-327n-1
    4. (2x2ay-b)3(2x2ay-b)3
    5. 23x-1·8x+142x-223x-1·8x+142x-2
    6. 62x·112x222x-1·32x62x·112x222x-1·32x
    Click here for the solution
  4. Simplify, without using a calculator:
    1. (-3)-3·(-3)2(-3)-4(-3)-3·(-3)2(-3)-4
    2. (3-1+2-1)-1(3-1+2-1)-1
    3. 9n-1·273-2n812-n9n-1·273-2n812-n
    4. 23n+2·8n-343n-223n+2·8n-343n-2
    Click here for the solution

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