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 a^m 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 =  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 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