In this page we have ** NCERT book Solutions for Class 8th Maths:Exponents** for
EXERCISE 1 . Hope you like them and do not forget to like , social share
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**Question 1 **

Evaluate

i) 3^{ -2}

ii) (- 4)^{ - 2}

iii) (1/2)^{-5 }

**Answer**

As we know that

b^{-n}= 1/b^{n}

(i) 3^{-2} = 1/ 3^{2} =1/9

(ii) (-4)^{-2} =1/(-4)^{2}= 1/6

(iii) (1/2)^{-5 } =1/ 2^{-5}

=2^{5}= 32

**Question 2**

Simplify and express the result in power notation with positive exponent.

**Answer **

(i)

= (-4)^{5} / (-4)^{8}

= (-4)^{5-8}

=1/ (-4)^{3}

(ii)

= 1/2^{3x2}

=1/2^{6}

iv )

= (3^{-7}/ 3^{-10}) × 3^{-5}

=3^{-7+10} × 3^{-5}

=3^{3} × 3^{-5}

=3^{-2}

=1/3^{2}

v) 2^{ - 3} × ( - 7)^{ - 3}

= (2× -7)^{-3}

= (-14)^{-3}

=1/ (-14)^{3}

**Question 3**

Find the value of

- (3
^{0}+ 4^{ - 1}) × 2^{2} - (2
^{ - 1}× 4^{ - 1}) ÷ 2^{ - 2} - (3
^{ - 1}+ 4^{ - 1}+ 5^{ - 1})^{0} -
**Answer**

- (3
^{0}+ 4^{ - 1}) × 2^{2}= (1+1/4) × 4

= 4+1=5

2. (2^{ - 1} × 4^{ - 1}) ÷ 2^{ - 2}

= [(1/2) × (1/4)] ÷ (1/4)

= (1/8) × (4) =1/2

3.

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

=4+9+16=29

4. (3^{ - 1} + 4^{ - 1} + 5^{ - 1})^{0}

=1 as a^{0}=1

**Question 4**

Evaluate

Answer

**Question 5**

Find the value of *m* for which 5* ^{m}* ÷ 5

**Answer **

5* ^{m}* ÷ 5

5^{m} × (1/5^{-3}) = 5^{5}

5^{m+3} =5^{5}

So m+3=5

m=2

**Question 6 **

Evaluate

**Answer **

**Question 7**

Simplify.

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