## First Law

If a is any non-zero rational number and m, n are natural numbers, then

$${ a }^{ m }$$ x $${ a }^{ n }$$ = $${ a }^{ m + n }$$
Generalised form of above law:

If a is a non-zero rational number and m, n, p are natural numbers, then,

$${ a }^{ m }$$ x $${ a }^{ n }$$ x $${ a }^{ p }$$ = $${ a }^{ m + n + p }$$

Illustration : Simplify and write the answer of each of the following in exponential form:

(i) $${ 4 }^{ 2 }$$ x $${ 4 }^{ 3 }$$        (ii) $${ 2 }^{ 2 }$$ x $${ 2 }^{ 3 }$$ x $${ 2 }^{ 4 }$$
(iii) $${ 6 }^{ x }$$ x $${ 6 }^{ 3 }$$     (iv) $${ (\frac { 3 }{ 2 } ) }^{ 3 }$$ x $${ (\frac { 3 }{ 2 } ) }^{ 6 }$$

Solution. Using first law of exponents, We have

(i) $${ 4 }^{ 2 }$$ x $${ 4 }^{ 3 }$$ = $${ 4 }^{ 2 + 3 }$$ = $${ 4 }^{ 5 }$$

(ii) $${ 2 }^{ 2 }$$ x $${ 2 }^{ 3 }$$ x $${ 2 }^{ 4 }$$ = $${ 2 }^{ 2 + 3 + 4 }$$ = $${ 2 }^{ 9 }$$

(iii) $${ 6 }^{ x }$$ x $${ 6 }^{ 3 }$$ = $${ 6 }^{ x + 3 }$$

(iv) $${ (\frac { 3 }{ 2 } ) }^{ 3 }$$ x $${ (\frac { 3 }{ 2 } ) }^{ 6 }$$= $${ (\frac { 3 }{ 2 } ) }^{ 6 + 3 }$$ = $${ (\frac { 3 }{ 2 } ) }^{ 9 }$$

## Second Law

If a is any non-zero rational number and m and n are natural numbers such that m > n, then

$${ a }^{ m }$$ $$\div$$ $${ a }^{ n }$$ = $${ a }^{ m – n }$$ or $$\frac { { a }^{ m } }{ { a }^{ n } }$$ = $${ a }^{ m – n }$$

Illustration : Simplify and write each of the following in exponential form:

(i) $${ 8 }^{ 6 }$$ $$\div$$ $${ 8 }^{ 3 }$$        (ii) $${ (-5) }^{ 10 }$$ ÷ $${ (-5) }^{ 4 }$$
(iii) $${ (\frac { -3 }{ 5 } ) }^{ 6 }$$ ÷ $${ (\frac { -3 }{ 5 } ) }^{ 3 }$$
Solution. Using second law of exponents, We have

(i) $${ 8 }^{ 6 }$$ $$\div$$ $${ 8 }^{ 3 }$$

= $$\frac { { 8 }^{ 6 } }{ { 8 }^{ 3 } }$$
=  $${ 8 }^{ 6 – 3 }$$
=  $${ 8 }^{ 3 }$$
(ii) $${ (-5) }^{ 10 }$$ ÷ $${ (-5 )}^{ 4 }$$

= $$\frac { { (-5) }^{ 10 } }{ {(-5) }^{ 4 } }$$

= $${ (-5) }^{ 10 – 4 }$$
= $${ (-5) }^{ 6 }$$
(iii) $${ (\frac { -3 }{ 5 } ) }^{ 6 }$$ ÷ $${ (\frac { -3 }{ 5 } ) }^{ 3 }$$

= $${ \frac { { (\frac { -3 }{ 5 } })^{ 6 } }{ ({ \frac { -3 }{ 5 } ) }^{ 3 } } }$$
= $${ (\frac { -3 }{ 5 } ) }^{ 6 – 3 }$$
= $${ (\frac { -3 }{ 5 } ) }^{ 3 }$$

## Third Law

If a is any rational number different from zero and m, n are natural numbers, then

$${ { (a }^{ m }) }^{ n }\quad =\quad { a }^{ m\quad \times \quad n }\quad =\quad { { (a }^{ n }) }^{ m }$$

Illustration: Simplify and write each of the following in exponential form:

(i) $${ { (3 }^{ 2 }) }^{ 4 }\quad$$    (ii) $${ { ((-2) }^{ 4 }) }^{ 2 }\quad$$
(iii) $${ { \{ (\frac { 3 }{ 5 } ) }^{ 3 }\} }^{ 4 }$$    (iv) $${ ({ 4 }^{ 2 }) }^{ 3 }\quad$$ x $$({ { 4}^{ 4 }) }^{ 2 }\quad$$
Solution. Using third law of exponents, We have

(i) $${ { (3 }^{ 2 }) }^{ 4 }\quad$$

= $${ 3 }^{ 2 \times 4 }$$
= $${ 3 }^{ 8 }$$
(ii) $${ { ((-2) }^{ 4 }) }^{ 2 }\quad$$

= $${ (-2) }^{ 4 \times 2 }$$

= $${ (-2) }^{ 8 }$$
(iii) $${ { \{ (\frac { 3 }{ 5 } ) }^{ 3 }\} }^{ 4 }$$

= $${ (\frac { 3 }{ 5 } ) }^{ 3 \times 4 }$$
= $${ (\frac { 3 }{ 5 } ) }^{ 12 }$$

(iv) $${ ({ 4 }^{ 2 }) }^{ 3 }\quad$$x $${ ({ 4}^{ 4 }) }^{ 2 }\quad$$

= $$({ 4 }^{ 2 \times 3})$$ x $$({ 4 }^{ 4 \times 2 })$$
= $${ 4 }^{ 6 }$$ x $${ 4 }^{ 8 }$$
= $${ 4 }^{ 6 + 8 }$$
= $${ 4 }^{ 14 }$$

## Fourth Law

If a, b are non-zero rational numbers and n is a natural number, then

$${ a }^{ n }$$ x $${ b }^{ n }$$ = $${ (ab) }^{ n }$$
Generalised form of above law:

If a, b, c are non-zero rational numbers and n is a natural number, then

$${ a }^{ n }$$ x $${ b }^{ n }$$ x $${ c }^{ n }$$ = $${(abc)}^{ n }$$

Illustration: Express each of the following products of powers as the exponent of a rational number:

(i) $${ 2 }^{ 4 }$$ x $${ 5 }^{ 4 }$$    (ii) $${(-3) }^{ 3 }$$ x $${ (-2) }^{ 3 }$$
(iii) $${ 3 }^{ 2 }$$ x $${ x }^{ 2 }$$ x $${ y }^{ 2 }$$    (iv) $${ (\frac { 3 }{ 2 } ) }^{ 2 }$$ x $${ (\frac { 2 }{ 5 } ) }^{ 2 }$$
Solution. We have,

(i) $${ 2 }^{ 4 }$$ x $${ 5 }^{ 4 }$$

= $${ (2 \times 5 )}^{ 4 }$$
= $${ 10 }^{ 4 }$$
(ii) $${(-3) }^{ 3 }$$ x $${ (-2) }^{ 3 }$$

= $${ ((-3) \times (-2) )}^{ 3 }$$

= $${ 6 }^{ 3 }$$

(iii) $${ 3 }^{ 2 }$$ x $${ x }^{ 2 }$$ x $${ y }^{ 2 }$$

= $${( 3 \times x \times y )}^{ 2 }$$
= $${ (3xy) }^{ 2 }$$

(iv) $${ (\frac { 3 }{ 2 } ) }^{ 2 }$$ x $${ (\frac { 2 }{ 5 } ) }^{ 2 }$$

= $${ (\frac { 3 }{ 2 } }\times \frac { 2 }{ 5 } )^{ 2 }$$
= $${ (\frac { 3 }{ 5 } ) }^{ 2 }$$

## Fifth Law

If a and b are non-zero rational numbers and n is a natural number, then

$$\frac { { a }^{ n } }{ { b }^{ n } }$$ = $${ (\frac { a }{ b } ) }^{ n }$$

Illustration: Write each of the following in the form $$\frac{p}{q}$$

(i) $${ (\frac { 2 }{ 5 } ) }^{ 3 }$$    (ii) $${ (\frac { -3 }{ 2 } ) }^{ 4 }$$
Solution. We have,

(i) $${ (\frac { 2 }{ 5 } ) }^{ 3 }$$

= $$\frac { { 2 }^{ 3 } }{ { 5 }^{ 3 } }$$
= $$\frac{2 \times 2 \times 2}{5 \times 5 \times 5 }$$
= $$\frac{8}{125}$$
(ii) $${ (\frac { -3 }{ 2 } ) }^{ 4 }$$

= $$\frac { { (-3) }^{ 4 } }{ { 2 }^{ 4 } }$$

= $$\frac{(-3) \times (-3) \times (-3) \times (-3)}{2 \times 2 \times 2 \times 2}$$
= $$\frac{81}{16}$$

## Illustrative Examples

Example 1: Use the laws of exponents to simplify the following :

(i) $${ [{ (2^{ 3 }) }^{ 4 }] }^{ 5 }$$    (ii) $$[{ { 3 }^{ 6 }\quad \div \quad { 3 }^{ 4 }] }^{ 3 }$$
(iii) $${ 81 }^{ -1 }$$ x $${ 3 }^{ 5 }$$    (iv) $${ (\frac { 2 }{ 3 } ) }^{ 0 }$$ + $${ (\frac { 2 }{ 3 } ) }^{ -2 }$$
Solution. We have,

(i) $${ [{ (2^{ 3 }) }^{ 4 }] }^{ 5 }$$

= $${ [{ 2^{ 3 \times 4 }}] }^{ 5 }$$
= $${ [{ 2^{ 12 }}] }^{ 5 }$$
= $${ [{ 2^{ 12 \times 5}}] }$$
= $${ 2 }^{ 60 }$$
(ii) $$[{ { 3 }^{ 6 }\quad \div \quad { 3 }^{ 4 }] }^{ 3 }$$

= $${ [ 3^{ 6 – 4 }]^3}$$

= $${ [{ 3^{ 2 }}] }^{ 3 }$$
= $${ 3 }^{ 2 \times 3 }$$
= $${ 3 }^{ 6 }$$

(iii) $${ 81 }^{ -1 }$$ x $${ 3 }^{ 5 }$$

= $${ { (3 }^{ 4 }) }^{ -1 }\quad$$ x $${ 3 }^{ 5 }$$
= $${ 3 }^{ -4 + 5 }$$
= $${ 3 }^{ 1 }$$
= 3

(iv) $${ (\frac { 2 }{ 3 } ) }^{ 0 }$$ + $${ (\frac { 2 }{ 3 } ) }^{ -2 }$$

= 1 + $${ \frac { 1 }{ ({ \frac { 2 }{ 3 } ) }^{ 2 } } }$$
= $${ \frac { 1 }{ \frac { { 2 }^{ 2 } }{ { 3 }^{ 2 } } } }$$
= $${ \frac { 1 }{ \frac { 4 }{ 9 } } }$$
= 1 + $$\frac{9}{4}$$
= $$\frac{13}{4}$$

Example 2: Simplify and write the answer in the exponential form:

(i) $${ ({ { 2 }^{ 5 }\quad \div \quad { 2 }^{ 8 }) }^{ 5 }\quad \times \quad { 2 }^{ -5 }\quad }$$    (ii) $${ (-4) }^{ 3 }$$ x $${ (5) }^{ -3 }$$ x $${ (-5) }^{ -3 }$$

Solution.

(i) We have,

$${ ({ { 2 }^{ 5 }\quad \div \quad { 2 }^{ 8 }) }^{ 5 }\quad \times \quad { 2 }^{ -5 }\quad }$$
= $${ (\frac { { 2 }^{ 5 } }{ { 2 }^{ 8 } } ) }^{ 5 }$$ x $${ (2) }^{ -5 }$$
= $${ { (2 }^{ 5 – 8 }) }^{ 5 }$$ x $${ (2) }^{ -5 }$$
= $${ { (2 }^{ -3 }) }^{ 5 }$$ x $${ (2) }^{ -5 }$$
= $${ (2) }^{ -3 \times 5 }$$ x $${ (2) }^{ -5 }$$
= $${ (2) }^{ -15 }$$ x $${ (2) }^{ -5 }$$
= $${ (2) }^{ -15-5 }$$ = $${ (2) }^{ -20 }$$

(ii) We have,

$${ (-4) }^{ 3 }$$ x $${ (5) }^{ -3 }$$ x $${ (-5) }^{ -3 }$$
= $${ [-4 \times 5 \times (-5)] }^{ -3 }$$
= $${ (100) }^{ -3 }$$
= $${ { (10 }^{ 2 }) }^{ -3 }\quad$$
= $${ (10) }^{ 2 \times -3 }$$ = $${ (10) }^{ -6 }$$

Example 3: Simplify

(i) $${ ({ { 4 }^{ -1 } \div { 3 }^{ -1 }) }^{ -2 }}$$    (ii) $${ ({ { 5 }^{ 3 } \times { 3 }^{ -1 }) }^{ -1 }}$$ ÷ $${ 6 }^{ -1 }$$

Solution.

(i) $${ ({ { 4 }^{ -1 } \div { 3 }^{ -1 }) }^{ -2 }}$$

=  $${ (\frac { 1 }{ 4 } \times \frac { 3 }{ 1 } ) }^{ -2 }$$

=  $${ (\frac { 3 }{ 4 })^{-2}}$$

= $${ (\frac { 4 }{ 3 })^{2}}$$

= $$\frac{16}{9}$$

(ii) $${ ({ { 5 }^{ -1 } \times { 3 }^{ -1 }) }^{ -1 }}$$ ÷ $${ 6 }^{ -1 }$$

= $${ (\frac { 1 }{ 5 } \times \frac { 1 }{ 3 } ) }^{ -1 }$$ ÷ $$\frac{1}{6}$$

= $${ (\frac { 1 }{ 15 })^ {-1} }$$ ÷ $$\frac{1}{6}$$

= $$\frac{15}{1}$$ ÷ $$\frac{1}{6}$$

= $$\frac{15}{1} \times \frac{6}{1}$$

= 90

Example 4: Find the value of m if $${ (\frac { 2 }{ 9 })^ {3} }$$ x $${ (\frac { 2 }{ 9 })^ {-6} }$$ = $${ (\frac { 2 }{ 9 })^ {2m-1} }$$

Solution.

$${ (\frac { 2 }{ 9 })^ {3} }$$ x $${ (\frac { 2 }{ 9 })^ {-6} }$$ = $${ (\frac { 2 }{ 9 })^ {2m-1} }$$

=>    $${ (\frac { 2 }{ 9 })^ {3+(-6)} }$$ = $${ (\frac { 2 }{ 9 })^ {2m-1} }$$

=>    $${ (\frac { 2 }{ 9 })^ {-3} }$$ = $${ (\frac { 2 }{ 9 })^ {2m-1} }$$

In an equation, when bases on both sidess are equal, their powers must also be equal.

Therefore, 2m – 1 = -3 or 2m = -3 + 1

=>    2m = -2

=>    m = $$\frac{-2}{2}$$ = –1

Example 5: What number should $${ (\frac { 2 }{ 3 })^ {-2} }$$ be multiplied so that the product is $${ (\frac { 4 }{ 27 })^ {-1} }$$ ?

Solution.

$$x$$ x $${ (\frac { 2 }{ 3 })^ {-2} }$$  = $${ (\frac { 4 }{ 27 })^ {-1} }$$

=>    $$x$$ x  $${ (\frac { 3 }{ 2 })^ {2} }$$ = $$\frac{27}{4}$$

=>    $$x$$ x $$\frac{9}{4}$$ = $$\frac{27}{4}$$

=>    $$x$$ = $$\frac{27}{4}$$ x $$\frac{4}{9}$$

=>    $$x$$ = 3.

Hence, the required number is 3.

Example 6: Simplify

$$\frac { { 12 }^{ 4 }\times { 9 }^{ 3 }\times 4 }{ { 6 }^{ 3 }\times { 8 }^{ 2 }\times 27 }$$
Solution.    We have,

$$\frac { { 12 }^{ 4 }\times { 9 }^{ 3 }\times 4 }{ { 6 }^{ 3 }\times { 8 }^{ 2 }\times 27 }$$
= $$\frac { ({ 2 }^{ 2 }\times 3)^{4} \times ({ 3 }^{ 2 }) ^ {3} \times {2}^{2} }{ { {2}^ {3}\times 3 }^{ 3 }\times ({ 2 }^{ 3 })^{2}\times {3}^{3}) }$$
= $$\frac { { ({ 2 }^{ 2 }) }^{ 4 }\times { 3 }^{ 4 }\times { ({ 3 }^{ 2 }) }^{ 3 }\times { 2 }^{ 2 } }{ ({ 2 }^{ 3 }\times { 3 }^{ 3 })\times { { (2 }^{ 3 }) }^{ 2 }\times { 3 }^{ 3 } }$$
= $$\frac { { 2 }^{ 8 }\times { 3 }^{ 4 }\times {3}^{6} \times {2}^{2} }{ { 2 }^{ 3 }\times { 3 }^{ 3 }\times {2}^{6} \times {3}^{3} }$$
= $$\frac { ({ 2 }^{ 8 }\times { 2 }^{ 2 } )\times ({3}^{6} \times {3}^{4}) }{ ({ 2 }^{ 3 }\times { 2 }^{ 6 }\times ({3}^{3} \times {3}^{3}) }$$
= $$\frac { { 2 }^{ 8+2 }\times { 3 }^{ 4+6 } }{ { 2 }^{ 3+6 }\times { 3 }^{ 3+3 } }$$
= $$\frac { { 2 }^{ 10 }\times { 3 }^{ 10 } }{ { 2 }^{ 9 }\times { 3 }^{ 6 } }$$
= $$\frac { { 2 }^{ 10 } }{ { 2 }^{ 9 } }$$ x $$\frac { { 3 }^{ 10 } }{ { 3 }^{ 6 } }$$
= $${ 2 }^{ 10-9 }$$ x $${ 3 }^{ 10-6 }$$
= $${ 2 }^{ 1 }$$ x $${ 3 }^{ 4 }$$
= 2 x 81 = 162

Example 7: Find the values of n in each of the following:

(i) $${ ({ 2 }^{ 2 }) }^{ n }$$ = $${ ({ 2 }^{ 3 }) }^{ 4 }$$    (ii) $${ 2 }^{ 5n }$$ ÷ $${ 2 }^{ n }$$ = $${ 2 }^{ 4 }$$

Solution.

(i) We have,

$${ ({ 2 }^{ 2 }) }^{ n }$$ = $${ ({ 2 }^{ 3 }) }^{ 4 }$$
=>    $${ 2 }^{ 2n }$$ = $${ 2 }^{ 3 \times 4 }$$

=>    $${ 2 }^{ 2n }$$ = $${ 2 }^{ 12 }$$

=>    2n = 12        [On equating the exponents]

=>    n = $$\frac{12}{2}$$ = 6

(ii) We have,

$${ 2 }^{ 5n }$$ ÷ $${ 2 }^{ n }$$ = $${ 2 }^{ 4 }$$
=>    $$\frac { { 2 }^{ 5n } }{ { 2 }^{ n } }$$ = $${ 2 }^{ 4 }$$

=>    $${ 2 }^{ 5n-n }$$ = $${ 2 }^{ 4 }$$    [since, $$\frac { { a }^{ m } }{ { a }^{ n } }$$ = $${ a }^{ m-n }$$ ]

=>    $${ 2 }^{ 4n }$$ = $${ 2 }^{ 4 }$$

=>     4n = 4         [On equating the exponents]

=>    n = $$\frac{4}{4}$$ = 1

Example 8: If $${25}^{n-1}$$ + 100 = $${5}^{2n-1}$$, find the value of n.

Solution.    We have,

$${25}^{n-1}$$ + 100 = $${5}^{2n-1}$$

=>    $${5}^{2n-1}$$ – $${5}^{2n-1}$$ = 100

=>    $$\frac { { 5 }^{ 2n } }{ { 5} }$$ – $$\frac { { 25 }^{ n } }{ { 25} }$$ = $${10}^{2}$$

=>    $$\frac { { 5 }^{ 2n } }{ { 5} }$$ – $$\frac { ({ 5 } ^{2})^{ n } }{ { 25} }$$ = $$({2 \times 5}) ^{2}$$

=>    $$\frac { { 5 }^{ 2n } }{ { 5} }$$ – $$\frac { { 5 }^{ 2n } }{ { 25} }$$ = $${2}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ x $$\frac{1}{5}$$ – $$\frac { { 5 }^{ 2n } }{ { 25} }$$ = $${2}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ x $$(\frac{1}{5})$$ – $$\frac { { 5 }^{ 2n } }{ { 25} }$$ = $${2}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ x $$(\frac{5 – 1}{25})$$ = $${2}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ x $$(\frac{4}{25})$$ = $${2}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ x $$\frac { { 2 }^{ 2 } }{ { 5 }^{ 2 } }$$ = $${2}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ x $${2}^{2}$$ = $${2}^{2} \times {5}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ = $$\frac { { 2 }^{ 2 } \times {5}^{2} \times {5}^{2} }{ { 2 }^{ 2 } }$$

=>    $${5}^{2n}$$ = $${2}^{2 – 2}$$ x $${5}^{2 + 2}$$

=>    $${5}^{2n}$$ = $${2}^{0}$$ x $${5}^{4}$$

=>    $${5}^{2n}$$ = $${5}^{4}$$

=>    2n = 4     [On equating the exponents]

=>    n = $$\frac{4}{2}$$ = 2