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**Q1. Using laws of exponents, simplify and write the answer in exponential form**

**(i) 2 ^{3}Ã—2^{4}Ã—2^{5}**

**(ii) 5 ^{12}Ã·5^{3}**

**(iii) (7 ^{2})^{3}**

**(iv) (3 ^{2})^{5}Ã·34**

**(v) 3 ^{7}Ã—2^{7}**

**(vi) (5 ^{21}Ã·5^{13})Ã—5^{7}**

**Sol:**

(i) 2^{3}Ã—2^{4}Ã—2^{5}

We know that, a^{m}+a^{n}+a^{p }= a^{m+n+p}

So, 2^{3}Ã—2^{4}Ã—2^{5} = 2^{3+4+5}

= 2^{12}

(ii) 5^{12}Ã·5^{3}

We know that, a^{m}Ã·a^{n }= a^{mâˆ’n}

So, 5^{12}Ã·5^{3} = 5^{12âˆ’3}

= 5^{9}

(iii) (7^{2})^{3}

We know that, (am)n=amn

So, (72)3 = 7(2)(3)

= 76

(iv) (3^{2})^{5}Ã·3^{4}

We know that, a^{m}Ã·a^{n }= a^{mâˆ’n} and (a^{m})^{n }= a^{mn}

So, (3^{2})^{5}Ã·3^{4} = 3^{10}Ã·3^{4}

= 3^{10âˆ’4}

= 3^{6}

(v) 3^{7}Ã—2^{7}

We know that, (a^{m}Ã—b^{m})=(aÃ—b)^{m}

So, 3^{7}Ã—2^{7} = (3Ã—2)^{7}

= 6^{7}

(vi) (5^{21}Ã·5^{13})Ã—5^{7}

We know that, a^{m}Ã·a^{n}=a^{mâˆ’n} and(a^{m}Ã—a^{n})=(a)^{m+n}

So, (5^{21}Ã·5^{13})Ã—5^{7} = (5^{21âˆ’13})Ã—5^{7}

= (5^{8})Ã—5^{7}

= 5^{8+7}

= 5^{15}

**Q2. Simplify and express each of the following in exponential form**

**Sol:**

**Q3. Simplify and express each of the following in exponential form**

**Sol:**

We know that,

We know that,

**Q4. Write 9 ****Ã—9 ****Ã—9 ****Ã—9 ****Ã—9 in exponential form with base 3**

**Sol:**

9 Ã—9 Ã—9 Ã—9 Ã—9 = (9)^{5} = (3^{2})^{5}

= 3^{10}

**Q5. Simplify and write each of the following in exponential form**

**Sol:**

**Q6. Simplify**

**Sol:**

**Q7. Find the values of n in each of the following**

**Sol:**

Equating the powers

= 2n + 3 = 11

= 2n = 11- 3

= 2n = 8

= n = 4

(ii) 9Ã—3^{n} = 3^{7}

= 3^{2}Ã—3^{n} = 3^{7}

= 3^{2+n }= 3^{7}

Equating the powers

= 2 + n = 7

= n = 7 â€“ 2

= n = 5

Equating the powers

= n + 5 = 5

= n = 0

Equating the powers

Equating the powers

= 4 + 5 = 2n + 1

= 2n + 1 = 9

= 2n = 8

= n = 4

= 3(2n â€“ 2) =2(2n â€“ 2)

= 6n â€“ 6 = 4n â€“ 4

= 6n â€“ 4n = 6 â€“ 4

= 2n = 2

= n = 1

**Q8.** ** find the value of n**

**Sol:**

On equating the coefficient

3n â€“ 15 = -3

3n = -3 + 15

3n = 12

n = 4