INDICES

FURTHER MATHEMATICS SSS1 FIRST TERM

WEEK 1

INDICES

Contents

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

 

Basic Concept of Laws of Indices

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

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

1.       a ^m x  a ⁿ =  a^{m + n}    --------------------Multiplication law

 

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

                  Or p³   x  p² =  p ^{3 + 2} =  p⁵

 

2.       a^m ÷ aⁿ   = a^{m - n}     ---------------------Division law

            Example:  p⁶  ÷  p⁴  =  p ^{6  -  4} = p²

 

3.       (a ^m )ⁿ  = a^mn   ----------------Power law

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

                               Or  p^{3 x 2} = p⁶

 

4.       a^m ÷ a^m  =  a^{m - m}  =  a⁰  = 1

 a^m ÷a^m = a^m/a^m = a^o = 1

 a⁰ = 1  ………………………Zero Index

    Note: Any number raised to the power of  zero is 1

 

             Example:  3^o = 1,   c^o = 1,    y^o = 1

 

5.         (ab)^m = a^mb^m    -------------Product power law

 e.g. (2xy)²  = 4x²y²  

 

6.       a ^{– m} = 1/a^m        ------------- Negative Index

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

 

7.       a^1/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 = (a^m) ^1/n =   (ⁿ√a)^m 

 

             Example: 27^2/3 = 3√27 = 3² = 9.

 

Evaluation

1.  27^5/3            2.   1000000000⁰        3. 2^x-1 x 2^2x+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) 625^3/8 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

 

Theory

       1.   16^3/2 x 8^2/3               2.  3x² x 4x³

                   32^1/5                                           6x⁷