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Exam Lessons Further Mathematics Mathematics

Indices: Basic Laws and Application of Indices

CONTENT

  • Basic Concept of Laws of Indices
  • Application of Laws of Indices

Basic Concept of Laws of Indices

A number of the form am where a is a real number, a is multiplied by itself m times,

The number a is called the base and the super script m is called the index (plural indices) or exponent.

1.         a m x  a n =  am + n    ——————–Multiplication law

             Example:  p3   x  p2  = ( p  x  p x  p)  x  (p  x  p)  =  p 5

                  Or p x  p2 =  p 3 + 2 =  p5

2.         am ÷ an   = am – n     ———————Division law

            Example:  p6  ÷  p=  p 6  –  4 = p2

3.         (a m )= amn   —————-Power law

            Example:  (p3)2 = p3 x p3 =  p 3 + 3 =  p6

                               Or  p3 x 2 = p6

4.         am ÷ am  =  am – m  =  a= 1

 am ÷am = am/am = ao = 1

 a0 = 1  ………………………Zero Index

    Note : Any number raised to power of  zero is 1

             Example:  3o = 1,   co = 1,    yo = 1

5.         (ab)m = amb  ————-Product power law

 e.g. (2xy)= 4x2y2  

6.         a – m = 1/am        ————- Negative Index

            Example:  2 -1 =  ½,    and   3 -2  =  1/3 2 = 1/9

7.         a1/n  = n√a  ————– Root power law

            Example :   9 ½ = √9 = 3

                                27 1/3 =3√27 = 3 ie (3)3 = 3

8.         a m/n = (a 1/n) m = (n√a)m   ———–Fraction Index

           or a m/n = (am) 1/n =   (n√a)m 

             Example: 272/3 = 3√27 = 32 = 9.

Evaluation

1.  275/3            2.   10000000000        3. 2x-1 x 22x+2

Application of Laws of Indices

Examples

Solve the following

(i)         32 3/5                      (ii)       343 2/3 (iii) 64 2/3   (iv) 0.001   (v) 14 0

Solution:

i)          32 3/5 = (32 1/5) 3 = (5√32) 3 

         = 2 3 = 8

ii)         343 2/3   =  (343 1/3 )2   =   (3√343)2

             = (7 3)1/3)2

                         = 72 = 49

iii)         64 2/3 = (64 1/3)2 = (4 3)1/3)2 = 4 2

iv)        (0.001)3 = (1/100)3 = (1/10)3)3 = (10 -3)3

= 10 -9 = 1/10 9

v)         14 0 = 1

General Evaluation

Simplify the following     (a)     216 4/3   (b) 25 1.5   (c) (0.00001)2   (d) 32 2/5    (e) 81 ¾  (f) 6253/8 x 25

Reading Assignment : Further Mathematics project book 1(New third edition).Chapter 2 pg.4 – 6

Weekend Assignment

1)         Evaluate 3 x  = 1/81                        (a) 4      (b) -4        (c) -2         (d) 2

2)         Simplify            2r5 X 9r3              (a) p2                        (b) 2p4        (c) P3         (d) 18r8

3)         Solve 3-y = 243                          (a) -5                       (b) 5           (c) 3                      (d) -3

4)         Solve 25-5n = 625                          (a) 1/5       (b) 2/5     (c) 1 1/5      (d) – 2/5

5)         Simplify (0.0001)2                         (a) 10-5      (b) 10 -3    (c) 10-8        (d) 10-10

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