MEI C1 Study Resources Core1 Indices 2 Indices

MEI Core 1 Indices Section 2: Working with indices Study plan Background You have dealt with squares and cubes for a long time and you will have met other powers at G.C.S.E. Indices (powers) are a useful way of simplifying expressions and writing large and small numbers in a more convenient form such as Standard Form that you met at G.C.S.E. The rules governing them need to be learnt thoroughly as you will need them in other parts of the course and possibly in other subjects such as Physics too.

Detailed work plan 1. Read the sections on pages 130 – 135 paying particular attention to the more difficult processes in examples 5.8 to 5.12. The algebraic examples will help you later when simplifying expressions. 2. You can see further examples using the Flash resources Laws of indices and Zero, negative and fractional indices. You can also look at the Indices video. 3. Exercise 5B Try at least half of the parts in questions 1, 3, 5, 6*, 7, 9, 11*. 4. For some extra practice try the Standard numeric indices puzzle, the Advanced numeric indices puzzle and the Algebraic indices puzzle.

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MEI Core 1 Indices Section 2: Working with indices Notes and Examples In this section you will learn to manipulate and simplify expressions involving indices, or powers. You will also meet negative and fractional indices. These notes contain subsections on  The rules of indices  Negative indices  Fractional indices  More difficult examples

The rules of indices Three rules of indices are: 1. 2. 3.

am  an = am+n am  an = am-n (am)n = amn You can investigate these rules and see why they work by trying them out with simple cases, writing the sums out in full: E.g., to demonstrate rule 3:

 2    2  2  2 3 2

2

  2  2  2   2  2  2  2 2 2 2 2 2  26  232 Try some for yourself.

The number being raised to a power (a in this case) is called the base. Note: You can only apply these rules to numbers involving the same base. So, for example, you cannot apply the rules of indices to 23  35 . Can you explain why?

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MEI C1 Indices Section 2 Notes and Examples Example 1 Simplify (i) 24  27

(ii) 39  34

(iii) (53 )6

(iv) 23  43 using the first rule

Solution (i) 24  27  247

 211 using the second rule 94

(ii) 3  3  3 9

4

 35 using the third rule

(iii) (53 )6  536

 518 (iv) 23  43  23  (22 )3

 2 2 3

At first sight this looks as if it cannot be simplified, as the bases are different. However, 4 can be written as a power of 2.

6

 23  6  29

You can see some similar examples using the Flash resource Laws of indices.

Negative indices There are two more rules, which follow from the three already introduced:

4. a n 

1 an

5. a0  1 for any value of a. Again, it’s worth experimenting with numbers to get a feel for how and why these rules work. e.g. 21  1 2 1  2  21  2  1  1 2 And from rule 3,

21  21  211  20  1 Try some for yourself.

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MEI C1 Indices Section 2 Notes and Examples Note that it might seem strange that a 0  1 for any value of a, but if this were not so, the other rules would be inconsistent. If you consider a graph of y  a x , for different values of a, you will see that it is perfectly natural that a 0  1 . Try this on your graphical calculator.

Example 2 Find, as fractions or whole numbers, (i) 24 (ii) 52 (iii) 30 Solution

1 1  4 2 16 1 1 (ii) 52  2  5 25 0 (iii) 3  1 (i)

24 

Fractional indices

a1n  n a or

Although these are equivalent, it is usually easier to use the first form, working out the root first so that you are dealing with smaller numbers.

amn  (a1n )m  (n a )m amn  (am )1n  n am

As before, try experimenting with numbers to get a feel for how and why these rules work. Example 3 Find, as fractions or whole numbers, 3

1

(ii) 9 2

(i) 8 3

 12

(iii) 25

(iv) 4

5 2

Solution 1 (i) 8 3  3 8  2 3

(ii) 9 2  ( 9)3  33  27

1 1 1  1  25 5 25 2 5 1 1 1 1 1   5 (iv) 4 2  5  1 5  5 2 2 4 (4 ) ( 4) 2 32  12

(iii) 25



You can see more examples like the ones above using the Flash resource Zero, negative and fractional indices. You can also look at the Indices video.

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MEI C1 Indices Section 2 Notes and Examples More difficult examples The next example shows how you can sometimes simplify quite complicated looking expressions involving different bases by splitting them up into factors.

Example 4 Simplify 65 2 

1 183 2 12

Solution

65 2  (2  3)5 2  25 2  35 2 1 1 1 1 1     1 2  21  31 2 12 4 3 2 3 3 2 3 2 18  9  23 2  (91 2 )3  23 2  33  23 2

So 65 2 

1 183 2  25 2  35 2  21  31 2  33  23 2 12  (25 2  21  23 2 )  (35 2  31 2  33 )  20  31  1 31  13

A common mistake when dealing with indices is to try to add terms with the Can you same base but a different index, such as 23  25 , by adding the indices. explain why This is wrong, but you can sometimes simplify expressions like this by taking this is wrong? out a common factor. This is shown in the example below.

Example 5 Simplify 25/ 2  21 2 . Solution 25/ 2  21 2  22  21 2  21 2

 4  21 2  21 2  5  21 2 5 2

For further practice in manipulating indices, there are three puzzles in which you need to match equivalent expressions to form a large hexagon. There is a numeric indices puzzle, an advanced numeric indices puzzle (more difficult examples) and an algebraic indices puzzle (in which the expressions to be manipulated are algebraic).

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Core 1 Indices Section 2: Working with indices Crucial points 1. Make sure you use the law of indices in appropriate situations Remember you cannot apply the laws of indices to the sum or difference of two expressions involving indices (although you may be able to simplify in another way.)

8 9

8

Wrong

a 2 + a5 = a7

Right

a 2 + a 5 = a 2 (1 + a 3 )

9

2. Look at the base Make sure that you only apply the first two laws of indices to expressions with the same base

8 9

Wrong

22 × 35 = 67

Right

2 2 × 25 = 2 7

8 9

3. Remember the value of a0 a0 is always 1, for any value of a.

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MEI Core 1 Indices Section 2: Working with Indices Exercise Do not use a calculator in this exercise. 1. Find: (i) 34 (iv) 60 (vii) 16−1 2 (x)

⎛1⎞ ⎜ ⎟ ⎝2⎠

(ii) 26 (v) 5−2 (viii) 85 3

−1

(iii) 41 2 (vi) 641 3 (ix) 36−3 2

(xi)

⎛ 25 ⎞ ⎜ ⎟ ⎝ 9 ⎠

−1 2

⎛ 27 ⎞ (xii) ⎜ ⎟ ⎝ 64 ⎠

−2 3

2. Simplify the following: (i)

311 × 3−4 ÷ 33

(ii)

( 25 ) × ( 2 7 )

(iv)

25 × 41 2 2

(v)

(3 )

3. Simplify the following: (i) 35 2 − 31 2 (ii)

3

5 32

−2

× 9−7 4

21 2 + 23 2 + 25 2

(iii)

56 55 × 53

(vi)

x4 3 x1 3 × x8 3

(iii)

y1 2 − y −1 2

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MEI Core 1 Indices Section 2: Working with Indices Solutions to Exercise 3 4 = 3 × 3 × 3 × 3 = 81 2 6 = 2 × 2 × 2 × 2 × 2 × 2 = 64 12 4 = 4=2 60 = 1 1 1 5 −2 = 2 = (v) 5 25 13 3 (vi) 64 = 64 = 4 1 1 −1 2 (vii) 16 = = 16 4

1. (i) (ii) (iii) (iv)

= ( 3 8 ) = 2 5 = 32 1 1 1 −3 2 36 = = 3 = 3 ( 36 ) 6 216

(viii) 8

(ix)

5

5 3

−1

(x)

⎛1⎞ ⎜ ⎟ ⎝2⎠

−1 2

(xi)

⎛ 25 ⎞ ⎜ ⎟ ⎝ 9 ⎠

⎛ 27 ⎞ (xii) ⎜ ⎟ ⎝ 64 ⎠

−2 3

2. (i)

= ( 2 −1 )

−1

= 21 = 2

9 ⎞1 / 2 = ⎜⎛ = ⎟ ⎝ 25 ⎠ ⎛ 64 ⎞ =⎜ ⎟ ⎝ 27 ⎠

9 3 = 25 5 2

2/3

2 ⎛ 3 64 ⎞ ⎛ 4 ⎞ = 16 =⎜ = ⎜ ⎟ ⎟ 9 ⎝3⎠ ⎝ 27 ⎠

3 11 × 3 −4 ÷ 3 3 = 3 11 − 4− 3 = 3 4 = 81

(ii)

(25 ) × (2 7 )

(iii)

56 1 1 = 5 6 − 5 − 3 = 5 −2 = 2 = 5 3 5 ×5 5 25

(iv)

25 × 4 2

(v)

( 35 )

(vi)

3

12

32

−2

= 2 15 × 2 −14 = 2 15 − 14 = 2 1 = 2

12

2 5 ×(2 2 ) = 2

× 9

−7 4

= ( 35 )

x43 x 13 83 = x ×x

25 × 21 = = 2 5 + 1 − 1 = 2 5 = 32 2

32

4 1 8 3−3−3

× ( 32 )

=

−7 4

x− = (x− 5 3

7

= 3 15 / 2 × 3 −7 / 2 = 3 2 − 2 = 3 4 = 81 15

5 3

)

1 2

= x −6

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MEI C1 Indices section 2 Exercise solns

3. (i)

3

5 2

−3

12

= 31 / 2 × 32 − 3

12

= 3 1 / 2(3 2 − 1) = 3 ×8 =8 3

(ii)

2

12

+2

32

+2

5 2

= 21 / 2 + 21 / 2 × 21 + 21 / 2 × 22 = 2 1 / 2(1 + 2 + 2 2 ) =7 2

(iii)

y 1 2 − y −1 2 = y −

1

y

=

y −1 y

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Indices Hexagonal Jigsaw The three diagrams on the following pages will ultimately fit together to form a large hexagon. Before you start, the three diagrams must be cut along the lines to make twenty-four equilateral triangles. For the triangles to be fitted together, you must find two expressions that are equivalent to one another. To build up the puzzle, place the edges on which equivalent expressions are written together, so that the triangles are joined along this edge. Begin by just finding pairs of matching expressions and placing them edge-to-edge. As you progress you will find that all of the pieces will eventually link up to form a large hexagon. By all means try this by yourself, but it is really designed to be a group activity. Working with other students will help to highlight any misconceptions you may have, as well as making the task more enjoyable. At first glance this may appear to be an easy undertaking, but you will find it takes quite a lot of thought and errors can be made very easily. Make sure you check each coupling, or you could be left with one or two pieces that appear to fit nowhere.

© Susan Whitehouse & MEI, 19/02/07

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64

2

− 49 ½

12

3

−8

27

1 3 2

(−

10

7

×2

5

÷2

9

7)

1 49

2

−1

5

2

0 1

13

×

1 5

−2

16

4 32

2

5

2 3 2

49

−½

− 1 7

−2

75

1 ½

9

25 3 2

×3

16 ½ × 2 2

7

9

½

6

7 (10 ½) 2

1

49

81 ¼

2

1 2

×5 4

9 −4 −1

1

−7 2

−3

36

16 8

1 3

2) − (

4

6

3)

4 9

(

0

6

100 −½

3

2

÷6

12 1

3/

6

13

2 11

(16 2 ) ¼

−½

6

×6 9

½

−4

−7

9 )3

÷1 21

1

(

2

½

−½

1 144

1 10

Indices Hexagonal Jigsaw The three diagrams on the following pages will ultimately fit together to form a large hexagon. Before you start, the three diagrams must be cut along the lines to make twenty-four equilateral triangles. For the triangles to be fitted together, you must find two expressions that are equivalent to one another. To build up the puzzle, place the edges on which equivalent expressions are written together, so that the triangles are joined along this edge. Begin by just finding pairs of matching expressions and placing them edge-to-edge. As you progress you will find that all of the pieces will eventually link up to form a large hexagon. By all means try this by yourself, but it is really designed to be a group activity. Working with other students will help to highlight any misconceptions you may have, as well as making the task more enjoyable. At first glance this may appear to be an easy undertaking, but you will find it takes quite a lot of thought and errors can be made very easily. Make sure you check each coupling, or you could be left with one or two pieces that appear to fit nowhere.

© Susan Whitehouse & MEI, 19/02/07

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49 3 / 2 × 7 −2

1

1 4

÷( 2

2 3

×4

1 49

4

(16 5 / 4 ÷ 8 1 / 3) × 2 5

36

3 −3 0

×2

−5

−1

2 − 1/3

1

3

32 27

3

8

2/

27 ½

) 1 8 (0.

6 125

3

1

125 27

27

−½

) 2

×4

3 16

1

2

×9

−5

/2

16 27

−¼

6

18 ½ × 2 ½

1

1 27

1 11

9 10 (3

−½

) 1 8 (0.

3

)

½

½

32

10 0 9

0

7 16

(12 1)

−½

(0.64) − ½

× 32 16 ¼

(7

)

−4 ½

36 ½

¼

( 6 ½) 4

8

×9

5

1/

5

3

5 ×2

1/

3

9 4

1

5 ×2

0

−1

5 4

1 −2

×2

½

9

4

−3

4

1 3

(81 2 ) ½

1 4 /2

/3

¼

¼

27

×3

×(

2

3 5

1

−3

1 16

−1

8

2 1

10 9 −3

25 9

2

9

8

5/

/2

81

3)

−4

Indices Hexagonal Jigsaw The three diagrams on the following pages will ultimately fit together to form a large hexagon. Before you start, the three diagrams must be cut along the lines to make twenty-four equilateral triangles. For the triangles to be fitted together, you must find two expressions that are equivalent to one another. To build up the puzzle, place the edges on which equivalent expressions are written together, so that the triangles are joined along this edge. Begin by just finding pairs of matching expressions and placing them edge-to-edge. As you progress you will find that all of the pieces will eventually link up to form a large hexagon. By all means try this by yourself, but it is really designed to be a group activity. Working with other students will help to highlight any misconceptions you may have, as well as making the task more enjoyable. At first glance this may appear to be an easy undertaking, but you will find it takes quite a lot of thought and errors can be made very easily. Make sure you check each coupling, or you could be left with one or two pieces that appear to fit nowhere.

© Susan Whitehouse & MEI, 19/02/07

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x

3/

×x

2

−2

−6

1

x

3/

x

2

3

6

x

10

x

1

x3

x5

x

4

x3

x

x× x4

−5

1/

−

x

x 20

x 12

1

x 1/2× x 3/2× x 8

3

x

9

(x

2

)

x

)

2

2

/ 11

(x

−4

x2

x

8

3

7

x

(

x)

14

x

−¼ x

x9 3

x

3

10

( x)

1

x

−3

x

5

×

x

x −1 0

x 2 × x 5/2÷ x

− 1/2

x

( x 2)7 × x

3

)

3

(x

6

1/

x 13

x

6

x6

x

x x

−2

−7

x

1 10

x

×x 7

−1

x 5

x 4

11

x

x

10

x½

−9 4

− 1/3

x −¼

x

−¾

x

x

(x

)

×

−3

1/3

½

x

−4 x

×

1

1/2

x

x

−¾

x8 5

2/

3

)

(x

4

− 1/3

x

x 1/3× x 4/3

−½

x

/2

−3

11

x

x

−6

x

x

5

−6

x

×x

x

1

3

x

x x

6

x3 1 −

x

2 1/

−9

÷x

/2

(x

2

)

4

MEI Core 1 Indices Section 2: Working with indices Multiple Choice Test Do not use a calculator in this test. 1) 34 × 32 = (a) 98 (c) 96 (e) I don’t know

(b) 38 (d) 36

2) 510 ÷ 52 = (a) 512 (c) 58 (e) I don’t know

(b) 55 (d) 520

3) (24 )3 = (a) 212 (c) 64 (e) I don’t know

(b) 27 (d) 83

4) 3−4 =

1 81 (c) 4 3 (e) I don’t know (a)

1 81 4 (d) − 3

(b) −

5) 16−1 4 = (a) -2 (c) − 12 (e) I don’t know

(b) 2 (d) 12

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MEI C1 Indices Section 2 MC test 6) 27 2 3 = 1 18 (c) 18 (e) I don’t know

(a)

⎛ 4 ⎞ 7) ⎜ ⎟ ⎝ 25 ⎠

1 9 (d) 9

(b)

−3 2

=

125 8 125 (c) – 8 (e) I don’t know (a)

8 125 8 (d) – 125

(b)

8) Simplify 83 × 61 2 ÷ 323 2 (a) 3 2 (c) 4 3 (e) I don’t know

(b) 2 3 (d) 4 2

9) Simplify 35 2 − 33 2 + 31 2 (a) 217 3

(b) 4 3

(c) 7 3 (e) I don’t know

(d) 13 3

10) Simplify

91 3 ×12−1 2 31 6 × 20

(a) 14 (c) 2 (e) I don’t know

(b) (d)

1 2

2

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MEI Core 1 Uncertainty / Indices Chapter assessment This chapter assessment covers the work in both chapter 4 and chapter 5. Do not use a calculator for this test. 1. Write the following in terms of the simplest possible surd. (i) 27 (ii) 288 (iii) 96

[6]

2. Simplify (i) 98 − 32 (ii)

(

75 × 10 × 24

)(

(iii) 1 + 2 3 − 2 2

)

[9]

3. Rationalise the denominators of the following and simplify as far as possible 2− 3 12 (i) (ii) [2+3] 3 6

(iii)

1 3−2

4. Find the values of (i) 641 3 (iv) 25−1 2

(iv)

1+ 2 3− 2

(ii)

2−5

(v)

93 2

[5]

(ii)

( a 3 )5 ( a 2 )5 2

[4]

[3+4]

(iii)

( 13 )

0

5. Simplify (i) x3 × x8 ÷ x5 6. Simplify (i) 323 2 × 85 × 2−5 2 (ii) 10−1 3 × 252 3 ÷ 25 3

[6]

7. Solve the following inequalities. (a) 2x + 3 < 1 – x (b) 3(y – 1) ≥ 5y – 8

[6]

8. Solve the following inequalities. (a) x² + 2x – 15 ≤ 0 (b) 2p² – 7p + 3 > 0 (c) z(2 – z) < z – 12

[12] Total 60 marks

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