Indices – SS1 Mathematics Lesson Note (First Term)

1. LAWS OF INDICES
2. Xa x Xb = Xa+b
3. Xa ÷ Xb = Xa-b
4. X0 = 1
5. X-a=  1

Examples:

Simplify

1. 105 X 104        2. a3 X a4         
2. m8 ÷ m5         4.  24×6 ÷ 8×4      5.   198 ÷ 198

Solutions

1. 105 X 104 = 105+4 =109

2. a3 X a4 = a3+4 =a7

3. m8 ÷m5 = m8-5 = m3

4. 24×6 ÷ 8×4 =      24×6 =    3×6-4   =3×2

5. 198 ÷ 198 = 198-8 = 190 =1

CLASSWORK

Simplify

1. 6 x Z0    
2. 4-3    
3. Z3 x (⅙)1  
4. r x r x r x r-5

1. PRODUCT OF INDICES

(Xa)b = Xaxb = Xab

Examples

Simplify:

1. (X2)3             2. (Y4)2            3.   (3-2)-3      4. (-3d3)2      5. a6(-a)-4

Solutions

1. (X2)3 = X2X3 = X6

2. (Y4)2 = Y4X2 = Y8

3. (3-2)-3 = 3-2 X -3=3+6

             =36 =3 X 3 X 3 X 3 X 3 X3

         =27 X 27

          = 729

4. (-3d3)2 = (-3)2 X (d3)2

       = -3 X -3d6 = 9d6

5. a6(-a)-4 =  a6 X   1

  (-a)4

               =             a6

               (-a) X(-a) X (-a) X(-a)

    =  a6

               a4

              = a6 – 4

              = a2

CLASSWORK

Simplify

1. (h4)-5        2.  (-4u2v)3     3.  (-x3)2÷ x4     4.  – (d2) ÷ d4 x –d     5.  (-c)2 X (c)4 ÷ (-c3)

FRACTIONAL INDICES

X (1/a) and X (a/b)

    X    is short for the square root of x

√X  X  √ X = X

Let √x = xp

Then 

Xp X xp =√ x X √ x= x1

By equating the indices

2p = 1   ,     P =½

Thus √x – x(½)  =  3 x

Similarly, 3  x is short for the cube root of x e.g3 8= 2. Since 2 X 2 X 2 =8

And 3√-27 = -3

 Since (-3 ) X (-3)  X (-3 )  = -27

3√x X 3√x X 3√x = x

i.exq X xq X xq = x1

x3q = 1

Equating the power 

3q=1

q= ⅓

thus3√x = x⅓

In general x1/a =a√x

Also  x2/3 = x2 X 1/3= (x2)1/3

=3√x2

OR

X2/3 = (x2 x 1/3)= (x1/3)2

= (3√x)2

In general 

Xa/b = b√xaor (b√x)a

Examples

Simplify

1. 8 – ⅔ = 1   2.  4⅙ X 4⅓   3. (16/81) – ¾.   4.√72a3b – (2/2b5b) – 6

Solutions:

1. 8 – ⅔ =    1

     =       (3√8)2  =  1     

    =        (2)2      =    4

2. 4⅙ X 4⅓ = 4⅙÷ ⅓

=  4 (3/6) = 4 (½)

=√4 = 2

3. (16/81) – ¾ =         1

       = ( 4√16/81)3 =    1

=   (2/3)3

= 1 ÷ (2/3)3

=1 ÷ (8/27) = 1 X (27/8) = 27/8

4. 72a2b-2    =     (72a3b-2)1/2

2b5b-6                2a5b-6

=  72 X a3 Xa-5 X b2 X b-6

     2

= √36a3-5 X b-2-(-6)

= √36a-2 X b-2+6

= √36a-2 X b4

= √36 X (a-2 X b4)1/2

=6 X a-2 X1/2 X b4x1/2

=6a-1 X b2 =6 X 1X b2

a

=6b2a

CLASSWORK 

Simplify:

1. (125)-1/3    2.   (18/32)-3/2      3.(3√4)1.5      4.64-5/6      5.   √1  9/16

SOLVING EQUATION WITH INDICES

Eg: Solve the following equations:

1. 2r-3 = -16
2. 5x = 40x – ½  

Solutions

1. 2r-3 = -16

Divide both sides by 2

2r-3 = -16

2          2

r-3 = -8

1   = -8

r3       1

-8r3 = 1 X 1

r3 = – 1

         8

Take the cube root of both sides

3√r3 = 3 – 1

             -8

r = -1

       2

1. 5x = 40x -½

5        5

x= 8x-1/2

x=  8 x  1

x1/2

Cross multiply 

xX x1/2 =8

x1 X x1/2 =8

x1+1/2 =8

x3/2 =8

i.e (√x)3= 8

raise both sides by power  2/3

(x3/2) X 2/3 = (8)2/3

X1= (3√8)2

X= (2)2

X= 4

4c-1 =64

Change both sides to the same base

4c-1 = 43

Equate the powers

c-1 = 3

c = 3 + 1

c = 4

CLASSWORK 

Solve the following equation

1. a⅔= 9      2.  2x(3)= 54

ASSIGNMENT 

Simplify the following:

1. 33 X 6-3 X 25                                                   (a) 0             (b) 1        (c) 2       (d)4           (e) 12

2. Calculate the value of (27/125)⅓ X (4/9) (½) 

(a)12/25      (b) 2/5    (c) 3/5   (d) 9/10    (e) 10/9

3. If 5p-3 = 8 X 5-2, find the value of p                   (a) 8/125    (b) 2/5   (c) 4/5   (d) 8/5       (e) 5/2

4. If x2 = 811½, x = ———-                                       (a) 3             (b) 9       (c)18     (d)27          (e) 54

5. Simplify (x1⅓)3 x 1 (a)1              (b) x1        (c) x3       (d) x0           (e)2

x4

6. Evaluate 9½ X 27⅔

7. Solve the following equations

1. Y-2 = 9       (b) (2s)½ = 9    (c) 2n-1 = 16