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

**(i) 3 ^{2} × 3^{4} × 3^{8}**

**Solution:**

We use the rule of multiply for the power with same base a^{m }x a^{n}= a^{m+n}

= (3)^{ 2+3+8}

= 3^{14}

**(ii) 6 ^{15} ÷ 6^{10}**

**Solution:**

By the rule of dividing for the powers with same base = a^{m} ÷ a^{n} = a^{m-n}

= (6)^{ 15-10}

= 6^{5}

**(iii) a ^{3} × a^{2}**

**Solution:**

= (a)^{ 3+2}

= a^{5}

**(iv) 7 ^{x} × 7^{2}**

**Solution:**

= (7)^{ x+2}

**(v) (5 ^{2})^{ 3} ÷ 5^{3}**

**Solution:**

(5^{2})^{ 3} can be written as = (5)^{ 2×3}

= 5^{6}

Now,

5^{6} ÷ 5^{3}

= (5)^{ 6-3}

= 5^{3}

**(vi) 2 ^{5} × 5^{5}**

**Solution:**

= (2 × 5)^{ 5}

= 10^{5}

**(vii) a ^{4} × b^{4}**

**Solution:**

= (a × b)^{ 4}

= (ab)^{4}

**(viii) (3 ^{4})^{ 3}**

**Solution:**

=(3^{4})^{ 3} can be written as = (3)^{ 4×3}

= 3^{12}

**(ix) (2 ^{20} ÷ 2^{15}) × 2^{3}**

**Solution:**

(2^{20} ÷ 2^{15}) can be simplified as,

= (2^{20 – 15})

= 2^{5}

Then,

2^{5} × 2^{3} can be simplified as,

= (2)^{5 + 3}

= 2^{8}

**(x) 8 ^{t} ÷ 8^{2}**

**Solution:**

= (8)^{t – 2}

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

(**i) (2 ^{3} × 3^{4} × 4)/ (3 × 32)**

**Solution:**

Factors of 4 = 2 × 2

= 2^{2}

Factors of 32 = 2 × 2 × 2 × 2 × 2

= 2^{5}

Then,

= (2^{3} × 3^{4} × 2^{2})/ (3 × 2^{5})

[we know that, a^{m }× a^{n} = a^{m+n}]

= (2^{5} × 3^{4}) / (3 × 2^{5})

= 2^{5-5} × 3^{4-1} … [by: a^{m }÷ a^{n} = a^{m-n}]

= 2^{0} × 3^{3}

= 1 × 3^{3}

= 3^{3}

**(ii) ((5 ^{2})^{3} × 5^{4}) ÷ 5^{7}**

**Solution:**

(5^{2})^{3} can be written as = (5)^{2×3}

= 5^{6}

Now,

= (5^{6} × 5^{4}) ÷ 5^{7}

= (5^{6+4}) ÷ 5^{7}

= 5^{10} ÷ 5^{7}

= 5^{10-7}

= 5^{3}

**(iii) 25 ^{4} ÷ 5^{3}**

**Solution:**

(25)^{4} can be written as = (5^{2})^{4}

(5^{2})^{4 }can be written as = (5)^{2×4}

= 5^{8}

Then,

= 5^{8} ÷ 5^{3}

= 5^{8-3}

= 5^{5}

**(iv) (3 × 7 ^{2} × 11^{8})/ (21 × 11^{3})**

**Solution:**

Factors of 21 = 7 × 3

Then,

= (3 × 7^{2} × 11^{8})/ (7 × 3 × 11^{3})

= 3^{1-1} × 7^{2-1} × 11^{8-3}

= 3^{0} × 7^{1} × 11^{5}

= 1 × 7 × 11^{5}

= 7 × 11^{5}

**(v) 3 ^{7}/ (3^{4} × 3^{3})**

**Solution:**

= 3^{7}/ (3^{4+3})

= 3^{7}/ 3^{7}

= 3^{7-7}

= 3^{0}

= 1

**(vi) 2 ^{0} + 3^{0} + 4^{0}**

**Solution:**

= 1 + 1 + 1

= 3

**(vii) 2 ^{0} × 3^{0} × 4^{0}**

**Solution:**

= 1 × 1 × 1

= 1

**(viii) (3 ^{0} + 2^{0}) × 5^{0}**

**Solution:**

= (1 + 1) × 1

= (2) × 1

= 2

**(ix) (2 ^{8} × a^{5})/ (4^{3} × a^{3})**

**Solution:**

(4)^{ 3} can be written as = (2^{2})^{ 3}

(5^{2})^{ 4} can be written as = (2)^{ 2×3}

= 2^{6}

Now,

= (2^{8} × a^{5})/ (2^{6} × a^{3})

= 2^{8-6} × a^{5-3}

= 2^{2} × a^{2}

= 2a^{2}

**(x) (a ^{5}/a^{3}) × a^{8}**

**Solution:**

= (a^{5-3}) × a^{8}

= a^{2} × a^{8}

= a^{10}

**(xi) (4 ^{5} × a^{8}b^{3})/ (4^{5} × a^{5}b^{2})**

**Solution:**

= 4^{5-5} × (a^{8-5} × b^{3-2})

= 4^{0} × (a^{3}b^{1})

= 1 × a^{3}b

= a^{3}b

**(xii) (2 ^{3} × 2)^{ 2}**

**Solution:**

= (2^{3+1})^{ 2}

= (2^{4})^{ 2}

= 2^{8}

**3. Say true or false and justify your answer:**

**(i) 10 × 10 ^{11} = 100^{11}**

**Solution:**

According to question:

Let us consider Left Hand Side (LHS) = 10 × 10^{11}

= 10^{1+11}

= 10^{12}

Now,

Right Hand Side (RHS) = 100^{11}

= (10 × 10)^{ 11}

= (10^{1+1})^{ 11}

= (10^{2})^{ 11}

= 10^{22}

Now,

By comparing LHS and RHS,

LHS ≠ RHS

Hence, the given statement is false.

**(ii) 2 ^{3} > 5^{2}**

**Solution:**

According t question:

Let us consider LHS = 2^{3}

2^{3} = 8

Now,

consider RHS = 5^{2}

5^{2} = 25

By comparing LHS and RHS,

LHS < RHS

2^{3} < 5^{2}

Hence, the given statement is false.

**(iii) 2 ^{3} × 3^{2} = 6^{5}**

**Solution:**

According to question:

Let us consider LHS = 2^{3} × 3^{2}

2^{3} × 3^{2} = 2 × 2 × 2 × 3 × 3

= 72

Now,

consider RHS = 6^{5}

6^{5 }= 7776

By comparing LHS and RHS,

72 ≠ 7776

LHS ≠ RHS

Hence, the given statement is false.

**(iv) 3 ^{0} = (1000)^{0}**

**Solution:**

Let us consider LHS = 3^{0 }= 1

Now,

consider RHS = 1000^{0} = 1

By comparing LHS and RHS,

LHS = RHS

3^{0} = 1000^{0}

Hence, the given statement is true.

**4. Express each of the following as a product of prime factors only in exponential form:**

**(i) 108 × 192**

**Solution:**

According to question:

The factors of 108 = 2 × 2 × 3 × 3 × 3

= 2^{2} × 3^{3}

The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

= 2^{6} × 3

Now,

= (2^{2} × 33) × (2^{6} × 3)

= 2^{2+6} × 3^{3+1}

= 2^{8} × 3^{4}

**(ii) 270**

**Solution:**

According to question:

The factors of 270 = 2 × 3 × 3 × 3 × 5

= 2 × 3^{3} × 5

**(iii) 729 × 64**

**Solution:**

According to question:

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3^{6}

The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2 = 2^{6}

Now,

= 3^{6} × 2^{6}

**(iv) 768**

**Solution:**

According to question:

The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= 2^{8} × 3

**5. Simplify:**

**(i) ((2 ^{5})^{2} × 7^{3})/ (8^{3} × 7)**

**Solution:**

We can write 8^{3} as = (2 × 2 × 2)^{3}

= (2^{3})^{3}

Now we have,

= ((2^{5})^{2} × 7^{3})/ ((2^{3})^{ 3} × 7)

= (2^{10}× 7^{3})/ ((2^{9 }× 7)

Now,

= (2^{10-9 }× 7^{3-1})

= 2 × 7^{2}

= 98

**(ii) (25 × 5 ^{2} × t^{8})/ (10^{3} × t^{4})**

**Solution:**

We can write 25 as = 5 × 5 = 5^{2}

10^{3} can be written as = (5 × 2)^{ 3}

= 5^{3} × 2^{3}

Now we have,

= (5^{2} × 5^{2} × t^{8})/ (5^{3} × 2^{3} × t^{4})

= (5^{2+2} × t^{8})/ (5^{3} × 2^{3} × t^{4})

Now,

= (5^{4} × t^{8})/ (5^{3} × 2^{3} × t^{4})

= (5^{4-3} × t^{8-4})/ 2^{3}

= (5t^{4})/ 8

**(iii) (3 ^{5} × 10^{5} × 25)/ (5^{7 }× 6^{5})**

**Solution:**

We can write 10^{5} as = (5 × 2)^{ 5}

= 5^{5} × 2^{5}

We can write 25 as = 5 × 5 = 5^{2}

We can write 6^{5} as = (2 × 3)^{ 5}

= 2^{5} × 3^{5}

Now we have,

= (3^{5} × 5^{5} × 2^{5} × 5^{2})/ (5^{7} × 2^{5} × 3^{5})

= (3^{5} × 5^{5+2} × 2^{5})/ (5^{7} × 2^{5} × 3^{5})

= (3^{5} × 5^{7} × 2^{5})/ (5^{5} × 2^{5} × 3^{5})

Now,

= (3^{5-5} × 5^{7-7} × 2^{5-5})

= (3^{0} × 5^{0} × 2^{0})

= 1 × 1 × 1

= 1

**👍👍👍**