# 1. Introduction

If a is any number and n any positive integer (whole number) then the product of a with itself n times, a1 × a2 × ... × an, is called a raised to the power n, and written an, i.e.:

an = a1 × a2 × ... × an

Example 1:

(a) 72 = 7 × 7 = 49
(b) 23 = 2 × 2 × 2 = 8
(c) 35 = 3 × 3 × 3 × 3 × 3 = 243

The following important rules apply to powers:

Rule 1 Rule 2 Rule 3 am × an = am + n am ÷ an = am − n (am)n = am × n a1 = a a0 = 1

We want these rules to be true for all positive values of a and all values of m and n. We shall first look at the simpler cases:

Example 2:

(a) 102 × 103 = (10 × 10) × (10 × 10 × 10) = 105 = 102+3
(b) 25 ÷ 23 = 32 ÷ 4 = 22 = 25−3
(c) (32)3 = (3 × 3)3 = (3 × 3) × (3 × 3) × (3 × 3) = 36 = 32×3
(d) From Rule 2, an+1 ÷ an = a1. Also:
an+1 ÷ an = a1 × a2 × a3 × ... × an+1/a1 × a2 × a3 × ... × an = a = a1
(e) We have an × a0 = an+0 = an = an×1. Thus a0 = 1.

Exercise 1: Simplify each of the following.

(a) 23 × 23

Solution: 23 × 23 = 2(3+3) = 26 = 64

(b) 315 ÷ 312

Solution: 315 ÷ 312 = 3(15−12) = 33 = 27

(c) (102)3

Solution: (102)3 = 10(2×3) = 106 = 1,000,000

Click on questions to reveal their solutions

# 2. Negative Powers

The question now arises as to what we mean by a negative power. To interpret this, note that

a2 ÷ a5 = a2/ a5 = a × a/a × a × a × a × a = 1/a3

If Rule 2 is to apply, then a2÷a5 = a2 − 5 = a−3. Thus a−3 = 1 ⁄ a3

The general rule is:

 a−n = 1 ⁄ an

Example 3:

(a)10−2 = 1102 = 1100
(b)3−1 = 131 = 13
(c)52 ÷ 54 = 52 − 4 = 5−2 = 1 ⁄ 52 = 125

Exercise 2: Write each of the following in the form ak, for some number k.

(a)23 × 2−5

Solution: 23 × 2−5 = 23−5 = 2−2, which is 14.

(b)35 ÷ 37

Solution: 35 ÷ 37 = 35−7 = 3−2, which is 19.

(c) (102)−3

Solution: (102)−3 = 10(2 × (−3)) = 10−6, which is 11,000,000.

Click on questions to reveal their solutions

# 3. Fractional Powers

If a is positive, then the square root of a is a number which, when muliplied by itself, gives a. Thus 3 is the square root of 9, since 32 = 9. We write 3 =  √9. Note that, by definition,  √a ×  √a = a. This gives us a way of interpreting a12 for, by Rule 1:

a12 × a12 = a(12 + 12) = a1 = a =  √a ×  √a

- so that a12 =  √a. The general rule is that, if a is a positive number and n is a positive integer, then:

 a1⁄n = n √a

- where n √a is the n th root of a. We can see this in general for, by Rule 3:

(a1n)n = a1n × n = a1 = a

Example 4:

(a) 10012 =  √100 = 10
(b) 813 = 3 √8 = 2
(c) 2753 = (2713)5 = 35 = 243

Note that in Example 4(c) we used Rule 3, i.e. (a1n)m = a1n × m = (a1m)n , so:

(a1n)m = amn = (a1m)n

Quiz 1: To which of the following does (85)13 simplify?

(a) 8 Incorrect - please try again!
(b) 16 Incorrect - please try again!
(c) 24 Incorrect - please try again!
(d) 32 Correct - well done!

Explanation: Using Rule 3, we have:

(85)13 = (813)5 = (2)5 = 32

# 4. Use of the Rules of Simplification

In this section we shall demonstrate the use of the rules of powers to simplify more complicated expressions.

Example 5:

(a)[(a−3)23]12

Beginning with the innermost bracket, we have, using Rule 3:

(a−3)23 = a−3 × 23 = a−2

Then:

[(a−3)23]12 = [a−2]12 = a−2 × 12 = a−1
(b)[(x−14)8] 23

Beginning again with the innermost bracket, and using Rule 3, we have:

(x−14)8 = x−14 × 8 = x−2

Now if we use Rule 3 again we have:

[x−2]23 = x−2 × 23 = x−43
(c)(x12)3 × (x−13)2

We have:

(x12)3 × (x−13)2 = x32 × x23

- using Rule 3. Now we use Rule 1:

x32 × x−23 = x3223 = x56
(d)(4 √x3)23 × (5 √x6)512

Starting with the first term:

4 √x3 = (x3)14 = x34

Thus:

(4 √x3)23 = (x34)23 = x34 × 23 = x24 = x12

Similarly:

5 √x6 = (x6)15 = x6 × 15 = x65

so that:

(5 √x6)512 = (x65)512 = x65 × 512 = x12

Finally we have:

(4 √x3)23 × (5 √x6)512 = x12 × x12 = x1 = x
(e)(a2/b3)13 × (b2/a3)12

The first term simplifies as follows.

(a2/b3)13 = (a2)13 / (b3)13 = a23 /b = a23b−1

Treating the second term:

(b2/a3)12 = (b2)12 / (a3)12 = b /a23 = ba−32

Thus:

(a2/b3)13 × (b2/a3)12 = a23b−1 × ba−32 = a2332 = a−56

# 5. Quiz on Powers

Simplify the expressions and choose the solutions from the options given.

1.(3 √a5)12 × 6 √a−5
(a)1
(b)a
(c)a512
(d)a56
2. (a3/b2)12 ÷ (b3/a2)−12
(a)a−12b−12
(b)a12b−12
(c)a−12b12
(d)a12b12
3.(4 √b3)16 × 9 √b−3 ÷ ( √b−7)17
(a)b524
(b)b−524
(c)b724
(d)b−724