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- Exponents And Power : Solution of NCERT Exercise 11.1

Exercise 1

1. Evaluate.

**(i) 3 ^{–2}**

**(ii) (– 4) ^{– 2}**

Answer: = 2Â²+3Â²+4Â² = 4+9+16 = 29

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

**Alternate Method:**

= -3^{4} x 5^{4} Ã·3^{4} (the power on 3 is even so -3^{4} = 3^{4})

=3^{0} x 5^{4}=625

(iv) (3^{–7} Ã·Answer: Using a^{m}Ã· a^{n} = a^{m-n}

(3^{–7} Ã· 3^{–10}) = 3^{-7+10} = 3^{3}

Hence, (3^{–7} Ã· 3^{–10}) Ã— 3^{–5}

=3^{3} x 3–5

Now, a^{m} x a^{n }= a^{m+n}

Hence, = 3^{3} x 3^{–5}

= 3^{3-5 }= 3^{-2}

(v) 2^{– 3} Ã— (–7)^{–3}

Answer: 2– 3 Ã— (–7)–3–3

Answer: 2– 3 Ã— (–7)–3

3. Find the value of.

(i) (3Â° + 4^{–1}) Ã— 2^{2}

Answer: (3Â° + 4^{–1}) Ã— 2^{2}

(ii) (2–1 Ã— 4–1) Ã· 2–2

(iii) (3^{–1} + 4^{–1} + 5^{–1})^{0}

Answer: aÂº=1

Hence, (3^{–1} + 4^{–1} + 5^{–1})^{0} = 1

**4. Evaluate **

(ii) (5^{–1} Ã— 2^{–1}) Ã— 6^{–1}

5. Find the value of m for which 5^{m} Ã· 5^{–3} = 5^{5}

Answer: a^{m} = a^{m-n}

Here, m-n = 5 and n = -3

So, m = 5+(-3)=2

6. Evaluate

Answer: (3-4)-1 = (-1)-1 =-1

7. Simplify.

(i) (25 x t^{-4}) Ã· (5^{-3} x 10 x t^{-8})

Answer: (25 x t^{-4}) Ã· (5^{-3} x 10 x t^{-8})

=(5Â² x t^{-4}) Ã· (5^{-3} x 5 x 2 x t^{-8})

= 5^{2+2} x t^{-4+8} Ã· 2

= 5^{4} x t^{4} Ã· 2

(ii) (3^{-5} x 10^{-5} x 125) Ã· (5^{-7} x 6^{-5})

Answer:

= (3^{-5} x 5^{-5} x 2^{-5} x 5^{3}) Ã· (5^{-7} x 6^{-5})

= (3^{-5} x 5^{-5+3} x 2^{-5}) Ã· (5^{-7} x 6^{-5})

= (3^{-5} x 5^{-2} x 2^{-5}) Ã· (5^{-7} x 3^{-5} x 2^{-5})

= (3^{-5+5} x 5^{-2+7} x 2^{-5+5})

= (3^{0} x 5^{5} x 2^{0})

=3125

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