## Law of Indices

Subject: Compulsory Maths

#### Overview

The laws of indices state a number of rules, which can be used to simplify expressions involving indices. Here, an index is used to write a product of numbers very compactly. The plural of index is indices.

##### Law of Indices

Indices is a number with the power. For example: am; a is called the base and m is the power. These laws only apply to expression with the same base.

Index help to write a product of numbers very compactly. Index help to show how many times to use the number in a multiplication. It is shown in the top right of the number in small number.

In this example: 4³ = 4x4x4 = 64

### Rule1: a° = 1

Any number, except 0, whose index is 0 is always equal to 1.

An example:

2° = 1

### Rule 2: a-m = $\frac{1}{a^m}$

An example:

2-3 = $\frac{1}{2^3}$ ( using a-m = $\frac{1}{a^m}$)

### Rule 3: amx an = am+n

In case of multiplication of same base, copy the base and add the indices.

An example:

3x 34 = 32+4 (using am x a= a m+n)

= 36

= 3 x 3 x 3 x 3 x 3 x 3

= 729

### Rule 4: am ÷ an = am-n

In case of division of same base, copy the base and subtract the indices.

An example:

w10 ÷ w6= w10-6 = w4

### Rule 5: ( am)n = amn

To raise an expression to the nth index, Copy the base and multiply the indices.

An example:

( x2)4 = x2x4 = x8

### Rule 6: a$\frac{m}{n}$ = ($\sqrt[n]{a}$)m

An example:

125$\frac{2}{3}$ = ($\sqrt[3]{125}$)2 = (5)2 = 25

##### Things to remember
• An indices is a number with the power.
• The laws of indices state a number of rules, which can be used  to simplify expressions involving indices.
• Any number, except 0, whose index is 0 is always equal to 1. (i.e. a° = 1)
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.

Solution,

(x3y)×(xy)×(x2y)

= x3× x× x2×y×y×y

=x3+1+2×b1+1+1

= x6y3

Solution:

35÷33

=$\frac{3^5}{3^3}$

=35-3

=32

=9

Solution:

12x7÷3x5

=$\frac{12x^7}{3x^5}$

=$\frac{3×4x^7-5}{3}$

=4x2

Solution:

$\frac{a^{m+n+2}×a^{m+n+2}}{a^{m+n}}$

=$\frac{a(^{(m+n+2)} \times\ a^{(m+n+2)}}{a^{m+n}}$

=a2(m+n+2)-(m+n)=a2m+2n+4-m-n

=am+n+4

Solution:

$\frac{-36a^8}{9a^5}$

=$\frac{-4×9a^{8-5}}{9}$

= -4a3

Solution:

(a2b)c×(ab2)c

=(a2)c.bc×ac(b2)c

=a2c.bc×ac.b2c

=a2c+c.bc+2c

=a3c.b3c

Solution:

$\frac{2^3×4^2}{8^2}$

=$\frac{2^3.(2^2)^2}{(2^3)^2}$

=$\frac{2^3.2^4}{2^6}$

=23+4-6

=27-6

=21

=2

Solution:

(a2b)×(ab)

=a2×a×b×b

=a2+1×b1+1

=a3b2

Solution:

$\frac{5^3×125^3}{25^3}$

=$\frac{5^3×(5^3)^3}{(5^2)^3}$

=$\frac{5^3×5^9}{5^6}$

=53+9-6

=512-6

=56

Solution:

$\frac{4^4×5^5}{25^2×16^2}$

=$\frac{4^4×5^5}{(5^2)^2×(4^2)^2}$

=$\frac{4^4×5^5}{5^4×4^4}$

=44-4×55-4

=40×51

=1×5

=5

Solution:

(-7p3)4

=(-7)4.(p3)4

=74.p3×4

=74p12

Solution:

(xy2)3×xy

=x3(y2)3×xy

=x3y2×3×xy

=x3.x.y6.y

=x3+1.y6+1

=x4y7

Solution:

(4x4)×(3x3)4=43(x4)3×34(x3)4

=43.x4×3.34.x3×4

=64×81.x12.x12

=43×34x12+12

=43×34 x24

Solution:

$\frac{(3p^2q)^2}{p^2q^2}$

$\frac{3^2(p^2)^2.q^2}{9p^2q^2}$

=$\frac{9p^4.q^2}{9p^2q^2}$

=p4-2q2-2

=p2.q0

=p2