Let us learn some rules to multiply and divide exponential numbers with the same base.

**Laws of Exponents**

Let us learn some rules to multiply and
divide exponential numbers with the same base.

__1. Multiplication of Numbers in Exponential form__

Let us calculate the value of 2^{3
}× 2^{2}

2^{3} ×2^{2}
=(2×2×2)×(2×2)

=2×2×2×2×2

= 2^{5}

=
2^{3}^{+}^{2}

We observe that the base of 2^{3}
and 2^{2} is the same 2 and the sum of the powers is 5. Now, let us consider
negative integers as the base.

(−3)^{3} ×
(−3)^{2}
=
[(−3)
×
(−3)
×
(−3)] × [(−3) × (−3)]

=(−3)
×
(−3)
×
(−3)
×
(−3)
×
(−3)

=(−3)^{5}

=(−3)^{3}^{+}^{2}

We observe that the base of (−3)^{3}
and (−3)^{2}
is the same as (−3) and the sum of the power is 5. Similarly,
*p* ^{4} × *p*^{2} =
( *p* × *p* × *p* ×
*p*) × ( *p* × *p*) =
*p*^{6} = *p*^{4} ^{+}^{2}

Now, for any non-zero integer ‘*a*’
and whole number ‘*m*’ and ‘*n*’, consider *a ^{m}* and

So, *a* * ^{m}* ×

* =a *×*
a *×*
a *×*
*...* *×*
a *(*m+n *times)* *=*a ^{m}
*

Therefore, *a* * ^{m}*
×

This is called Product Rule of exponents.

**Try these**

**Simplify and write the following in exponential form.**

**1. 2 ^{3} **

*2. p *^{2}* ***×*** p*^{4 }**= p^{6}**

*3. x *^{6}* ***×*** x*^{4}** = x^{10}**

**4. 3 ^{1}**

**5. (****−****1) ^{2} **

__Example 3.7 __

Simplify using Product Rule
of exponents.

(i) 5^{7} ×
5^{3}

(ii) 3^{3} ×
3^{2} × 3^{4 }

(iii) 25 ×
32 ×
625 ×
64

**Solution**

(i)_{ }5^{7} ×
5^{3} = 5^{7}^{+}^{3}

[since, *a* * ^{m}*
×

=
5^{10}

(ii) 3^{3} ×3^{2}
×3^{4}
=3^{3}^{+}^{2} ×3^{4} =3^{5
}× 3^{4}

=
3^{5}^{+}^{4} = 3^{9}

iii) 25×32×625×64
=(5×5)×(2×2×2×2×2)
× (5×5×5×5)×(2×2×2×2×2×2)

=5^{2} ×2^{5}
×5^{4}
×2^{6}

= (5^{2} × 5^{4}) × (2^{5} × 2^{6})
[grouping exponential numbers with the same base]

=
5^{2}^{+}^{4 }×
2^{5}^{+}^{6} = 5^{6 }×
2^{11}

__2. Division of Numbers in Exponential form__

Let us calculate the value of 2^{5}
÷
2^{2}

**Note**

Can you find the value
of *a *^{2}* *×* a*^{0}* *?

By Product Rule,

*a *^{2}* *×* a*^{0}* *=* a*^{2}* *^{+}^{0}

*a *^{2}* *×* a*^{0}* *=* a*^{2}

*a*^{0} = *a*^{2} / *a*^{2 }= 1

[dividing by *a*^{2}
on both sides]

Therefore, *a*^{0}
= 1 , *a* ≠ 0 .

We observe that the base of 2^{5}
and 2^{2} is the same ‘2’ and the difference of powers is 3. Now, let us
consider negative integers as the base.

Consider (−5)^{3}
÷
(−5)^{2}

^{}

(−5)^{3} / (−5)^{2
}=
[(−5)
×
(−5)
×
(−5)]
/ [(−5)
×
(−5)]

= (−5)^{1} = (−5)^{3}^{−}^{2}

We observe that the base of (–5)^{3}
and (–5)^{2} is the same as (–5) and the difference of the power is 1.

Thus, we can observe that for any non-zero
integer ‘*a*’ and for whole numbers ‘*m*’ and ‘*n*’, consider *a ^{m}*
and

That is, *a ^{m}
*=

*a ^{m}
/ a^{n} =* [

Therefore, a^{m }/ a^{n}
= a^{m −n}

This is called Quotient Rule of exponents.

__Example 3.8 __

Simplify using quotient rule
of exponents.

**Solution**

**Think**

What is Half of 2^{10}
? Ragu claims the answer is 2^{5} . Is he correct ? Discuss.

**Try these**

Simply the following.

**1. 23 ^{5} **

**2. 11 ^{6} ÷ 11^{3}**

**3. (−5) ^{3} ÷ (−5)^{2}**

**4. 7 ^{3} ÷7^{3}**

**5. 15 ^{4} ÷ 15**

__3. Power of Exponential form__

Let us find the value of (2^{2})^{5}

(
2^{2})^{5} = 2^{2} ×
2^{2} × 2^{2} × 2^{2} ×
2^{2} = 2^{2} ^{+} ^{2} ^{+}
^{2} ^{+} ^{2} ^{+}^{2} (By Product rule)

2^{10} =
2^{2}^{×}^{5}

Similarly, (3^{3})^{4} =3^{3} ×3^{3}
×3^{3}
×3^{3
}

=
3^{3}^{+}^{3}^{+}^{3}^{+}^{3
}=
3^{12 }= 3^{3}^{×}^{4}

(5^{6}
)^{2} = 5^{6} ×
5^{6} = 5^{6}^{+}^{6} = 5^{12} =
5^{6}^{×}^{2}

In general, for any non-zero integer
‘*a*’ and whole number ‘*m*’ and ‘*n*’,

(*
a ^{m} *)

*= a ^{m }*

*= _{ }a^{m }*

Hence, (* a ^{m} *)

This is called Power Rule of exponents.

**Try these**

**Simplify and write the following in exponent form.**

**1. (****3 ^{2}**

**2. [(****−****5****)**^{3}**] ^{2}**

**3. (****20 ^{6}**

**4. (****10 ^{3}**

__Example 3.9 __

Simplify using power rule
of exponents.

(i) ( 8^{3} )^{4}

(ii) (11^{5} )^{2}

(iii) ( 2^{6} )^{2} × ( 2^{4} )^{7}

**Solution**

(i) (8^{3})^{4} = 8^{3}^{×}^{4} = 8^{12 }[since ( *a* * ^{m}* )

(ii) (11^{5}
)^{2} = 11^{5×2} = 11^{10}
[since ( *a* * ^{m}* )

(iii) (2^{6}
)^{2} × (2^{4})^{7} = 2^{6}^{×}^{2} × 2^{4}^{×}^{7
}[since ( *a* * ^{m}* )

= 2^{12} × 2^{28}

=
2^{12 }+2^{28} = 2^{40}

[since (*a ^{m}*
×

**Think**

We learnt that 2^{2}
= 2 × 2 .

What is the value of 2^{22}
? Discuss.

__4. Exponent Numbers with Different Base and Same Power__

1. To understand
the multiplication of exponent numbers with different base and same powers, let
us consider the following example,

10^{5} =
10 ×10 ×10 ×10 ×10

= (2×5)×(2×5)×(2×5)×(2×5)×(2×5)

= (2×2×2×2×2)×(5×5×5×5×5)

10^{5} =2^{5}
×5^{5}

But we know that, 10 = 2 × 5. Hence 10^{5}
= (2 × 5)^{5} = 2^{5} × 5^{5} .

In general, for any non-zero integers
‘*a*’ and ‘*b*’ and for whole number ‘*m*’ (
*m* > 0),

*a ^{m} *×

= (*a* × *b*) × (*a* × *b*) × (*a* × *b*) × ... × (*a* × *b*) (*m* times) = (*a* × *b*)^{m}

Therefore, *a* * ^{m}*
×

2. To understand the division of exponent
numbers with different base and same powers, let us consider the follwing example,

10^{5} =
10 ×10 ×10 ×10 × 10

= (20/2) × (20/2) ×(20/2) ×(20/2) ×(20/2)

= [20×20×20×20×20] / [2×2×2×2×2]

Therefore, 10^{5} = 20^{5} / 2^{5} . But we know that,
10 = (20/2).

Hence, 10^{5} = (20/2)^{5}
= 20^{5} / 2^{5}.

Hence, for any two non-zero integers
‘*a*’ and ‘*b*’ and a whole number ‘*m*’ (
*m* > 0),

(*a*/*b*)* ^{m}* = (

= [ *a × a × a × ... × a* (*m* times)] / [ *b × b × b × ... × b* (*m*
times) ] = *a ^{m}* /

Therefore, .

**( a/b)^{m} = a^{m}
/ b^{m}**

^{}

**Try these**

**1.**** ****Express the following exponent
numbers using a^{m} **

(i) 5^{2} ×3^{2}** = (5 × 3) ^{2
}= 15^{2}**

(ii)* x *^{3} × y^{3}** = ( xy)^{3}**

(iii) 7^{4} ×8^{4}** = (7 × 8) ^{4}
= 56^{4}**

**2. Simplify the following exponent numbers by using ( a/b)^{m}
= a^{m} / b^{m} ^{}**

(i) 5^{3} ÷2^{3}^{ }= 5^{3} / 2^{3 }= (5 / 2)^{3}

(ii) (−2)^{4} ÷ 3^{4}** = ( –2) ^{4}
/ 3^{4} = (– 2 / 3)^{4}**

(iii) 8^{6} ÷5^{6}^{ }= 8^{6 }/ 5^{6} = (8 / 5)^{6}

(iv) 6^{3} ÷ (−7)^{3}^{ }= 6^{3 }/(–7)^{3 }= (6 / –7)^{3}

__Example 3.10 __

Simplify by using the law
of exponents.

(i) 7^{6} ×
3^{6}

(ii) 4^{3} ×
2^{3} × 5^{3}

(iii) 72^{5} ÷
9^{5}

(iv) 6^{13} ×
48^{13} ÷ 12^{13}

**Solution**

(i) 7^{6} ×3^{6}
= (7×3)^{6}
=21^{6
} [Since,
*a ^{m}* ×

(ii) 4^{3} ×2^{3}
×5^{3}
=(4×2×5)^{3}
=
40^{3 } [Rule
extended for 3 numbers]^{}

(iii) 72^{5} ÷ 9^{5}
= (72÷9)^{5} = 8^{5} [Since, *a ^{m}*
/

(iv) 6^{13} ×
48^{13} ÷ 12^{13} =
6^{13} × (48^{13} ÷
12^{13} ) [BIDMAS]

= 6^{13} ×
(48/12)^{13}^{ } [Since, *a* * ^{m}*
/

= 6^{13} ×
4^{13}

= (6 × 4)^{13 }[Since,
*a* * ^{m}* ×

= (24)^{13}

1. All the 10 digits appear
once in the expansion of 32043^{2}. That is, the value of 32043^{2}
= 1026753849 .

2. There are beautiful
equations with same exponent consecutive natural numbers as the base.

3^{2}+4^{2} =5^{2}

10^{2} + 11^{2} + 12^{2} = 13^{2} + 14^{2}

**Activity**^{}

**Finding the pair**

Divide the classroom into
two groups. Each group will have a set of cards. Each member of Group 1 has to pair
with one suitable member of Group 2 by stating the reason.

This activity can be extended till all the children in the class are familiarise with the laws of exponents.

_{ }

Tags : Algebra | Term 2 Chapter 3 | 7th Maths , 7th Maths : Term 2 Unit 3 : Algebra

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