Exercise 13.2 Page: 260

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

(i) 3² × 3⁴ × 3⁸

Solution:-

By the rule of multiplying the powers with same base = a^m × aⁿ = a^{m + n}

Then,

= (3)^{2 + 4 + 8}

= 3¹⁴

(ii) 6¹⁵ ÷ 6¹⁰

Solution:-

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

Then,

= (6)^{15 – 10}

= 6⁵

(iii) a³ × a²

Solution:-

By the rule of multiplying the powers with same base = a^m × aⁿ = a^{m + n}

Then,

= (a)^{3 + 2}

= a⁵

(iv) 7^x × 7²

Solution:-

By the rule of multiplying the powers with same base = a^m × aⁿ = a^{m + n}

Then,

= (7)^{x + 2}

(v) (5²)³ ÷ 5³

Solution:-

By the rule of taking power of as power = (a^m)ⁿ = a^mn

(5²)³ can be written as = (5)^{2 × 3}

= 5⁶

Now, 5⁶ ÷ 5³

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

Then,

= (5)^{6 – 3}

= 5³

(vi) 2⁵ × 5⁵

Solution:-

By the rule of multiplying the powers with same exponents = a^m × b^m = ab^m

Then,

= (2 × 5)⁵

= 10⁵

(vii) a⁴ × b⁴

Solution:-

By the rule of multiplying the powers with same exponents = a^m × b^m = ab^m

Then,

= (a × b)⁴

= ab⁴

(viii) (3⁴)³

Solution:-

By the rule of taking power of as power = (a^m)ⁿ = a^mn

(3⁴)³ can be written as = (3)^{4 × 3}

= 3¹²

(ix) (2²⁰ ÷ 2¹⁵) × 2³

Solution:-

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

(2²⁰ ÷ 2¹⁵) can be simplified as,

= (2)^{20 – 15}

= 2⁵

Then,

By the rule of multiplying the powers with same base = a^m × aⁿ = a^{m + n}

2⁵ × 2³ can be simplified as,

= (2)^{5 + 3}

= 2⁸

(x) 8^t ÷ 8²

Solution:-

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

Then,

= (8)^{t – 2}

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

(i) (2³ × 3⁴ × 4)/ (3 × 32)

Solution:-

Factors of 32 = 2 × 2 × 2 × 2 × 2

= 2⁵

Factors of 4 = 2 × 2

= 2²

Then,

= (2³ × 3⁴ × 2²)/ (3 × 2⁵)

= (2^{3 + 2} × 3⁴) / (3 × 2⁵) … [∵a^m × aⁿ = a^{m + n}]

= (2⁵ × 3⁴) / (3 × 2⁵)

= 2^{5 – 5} × 3^{4 – 1} … [∵a^m ÷ aⁿ = a^{m – n}]

= 2⁰ × 3³

= 1 × 3³

= 3³

(ii) ((5²)³ × 5⁴) ÷ 5⁷

Solution:-

(5²)³ can be written as = (5)^{2 × 3} … [∵(a^m)ⁿ = a^mn]

= 5⁶

Then,

= (5⁶ × 5⁴) ÷ 5⁷

= (5^{6 + 4}) ÷ 5⁷ … [∵a^m × aⁿ = a^{m + n}]

= 5¹⁰ ÷ 5⁷

= 5^{10 – 7} … [∵a^m ÷ aⁿ = a^{m – n}]

= 5³

(iii) 25⁴ ÷ 5³

Solution:-

(25)⁴ can be written as = (5 × 5)⁴

= (5²)⁴

(5²)⁴ can be written as = (5)^{2 × 4} … [∵(a^m)ⁿ = a^mn]

= 5⁸

Then,

= 5⁸ ÷ 5³

= 5^{8 – 3} … [∵a^m ÷ aⁿ = a^{m – n}]

= 5⁵

(iv) (3 × 7² × 11⁸)/ (21 × 11³)

Solution:-

Factors of 21 = 7 × 3

Then,

= (3 × 7² × 11⁸)/ (7 × 3 × 11³)

= 3^1-1 × 7^2-1 × 11^{8 – 3}

= 3⁰ × 7 × 11⁵

= 1 × 7 × 11⁵

= 7 × 11⁵

(v) 3⁷/ (3⁴ × 3³)

Solution:-

= 3⁷/ (3⁴⁺³) … [∵a^m × aⁿ = a^{m + n}]

= 3⁷/ 3⁷

= 3^{7 – 7} … [∵a^m ÷ aⁿ = a^{m – n}]

= 3⁰

= 1

(vi) 2⁰ + 3⁰ + 4⁰

Solution:-

= 1 + 1 + 1

= 3

(vii) 2⁰ × 3⁰ × 4⁰

Solution:-

= 1 × 1 × 1

= 1

(viii) (3⁰ + 2⁰) × 5⁰

Solution:-

= (1 + 1) × 1

= (2) × 1

= 2

(ix) (2⁸ × a⁵)/ (4³ × a³)

Solution:-

(4)³ can be written as = (2 × 2)³

= (2²)³

(5²)⁴ can be written as = (2)^{2 × 3} … [∵(a^m)ⁿ = a^mn]

= 2⁶

Then,

= (2⁸ × a⁵)/ (2⁶ × a³)

= 2^{8 – 6} × a^{5 – 3} … [∵a^m ÷ aⁿ = a^{m – n}]

= 2² × a²

= 2a² … [∵(a^m)ⁿ = a^mn]

(x) (a⁵/a³) × a⁸

Solution:-

= (a^{5 –} 3) × a⁸ … [∵a^m ÷ aⁿ = a^{m – n}]

= a² × a⁸

= a^{2 + 8} … [∵a^m × aⁿ = a^{m + n}]

= a¹⁰

(xi) (4⁵ × a⁸b³)/ (4⁵ × a⁵b²)

Solution:-

= 4^{5 – 5} × (a^{8 – 5} × b^{3 – 2}) … [∵a^m ÷ aⁿ = a^{m – n}]

= 4⁰ × (a³b)

= 1 × a³b

= a³b

(xii) (2³ × 2)²

Solution:-

= (2^{3 + 1})² … [∵a^m × aⁿ = a^{m + n}]

= (2⁴)²

(2⁴)² can be written as = (2)^{4 × 2} … [∵(a^m)ⁿ = a^mn]

= 2⁸

3. Say true or false and justify your answer:

(i) 10 × 10¹¹ = 100¹¹

Solution:-

Let us consider Left Hand Side (LHS) = 10 × 10¹¹

= 10^{1 + 11} … [∵a^m × aⁿ = a^{m + n}]

= 10¹²

Now, consider Right Hand Side (RHS) = 100¹¹

= (10 × 10)¹¹

= (10^{1 + 1})¹¹

= (10²)¹¹

= (10)^{2 × 11} … [∵(a^m)ⁿ = a^mn]

= 10²²

By comparing LHS and RHS,

LHS ≠ RHS

Hence, the given statement is false.

(ii) 2³ > 5²

Solution:-

Let us consider LHS = 2³

Expansion of 2³ = 2 × 2 × 2

= 8

Now, consider RHS = 5²

Expansion of 5² = 5 × 5

= 25

By comparing LHS and RHS,

LHS < RHS

2³ < 5²

Hence, the given statement is false.

(iii) 2³ × 3² = 6⁵

Solution:-

Let us consider LHS = 2³ × 3²

Expansion of 2³ × 3²= 2 × 2 × 2 × 3 × 3

= 72

Now, consider RHS = 6⁵

Expansion of 6⁵ = 6 × 6 × 6 × 6 × 6

= 7776

By comparing LHS and RHS,

72 ≠ 7776

LHS ≠ RHS

Hence, the given statement is false.

(iv) 3⁰ = (1000)⁰

Solution:-

Let us consider LHS = 3⁰

= 1

Now, consider RHS = 1000⁰

= 1

By comparing LHS and RHS,

LHS = RHS

3⁰ = 1000⁰

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:-

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

= 2² × 3³

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

= 2⁶ × 3

Then,

= (2² × 3³) × (2⁶ × 3)

= 2^{2 + 6} × 3^{3 + 1} … [∵a^m × aⁿ = a^{m + n}]

= 2⁸ × 3⁴

(ii) 270

Solution:-

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

= 2 × 3³ × 5

(iii) 729 × 64

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

= 3⁶

The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2

= 2⁶

Then,

= (3⁶ × 2⁶)

= 3⁶ × 2⁶

(iv) 768

Solution:-

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

= 2⁸ × 3

5. Simplify:

(i) ((2⁵)² × 7³)/ (8³ × 7)

Solution:-

8³ can be written as = (2 × 2 × 2)³

= (2³)³

We have,

= ((2⁵)² × 7³)/ ((2³)³ × 7)

= (2^{5 × 2} × 7³)/ ((2^{3 × 3} × 7) … [∵(a^m)ⁿ = a^mn]

= (2¹⁰ × 7³)/ (2⁹ × 7)

= (2^{10 – 9} × 7^{3 – 1}) … [∵a^m ÷ aⁿ = a^{m – n}]

= 2 × 7²

= 2 × 7 × 7

= 98

(ii) (25 × 5² × t⁸)/ (10³ × t⁴)

Solution:-

25 can be written as = 5 × 5

= 5²

10³ can be written as = 10³

= (5 × 2)³

= 5³ × 2³

We have,

= (5² × 5² × t⁸)/ (5³ × 2³ × t⁴)

= (5^{2 + 2} × t⁸)/ (5³ × 2³ × t⁴) … [∵a^m × aⁿ = a^{m + n}]

= (5⁴ × t⁸)/ (5³ × 2³ × t⁴)

= (5^{4 – 3} × t^{8 – 4})/ 2³ … [∵a^m ÷ aⁿ = a^{m – n}]

= (5 × t⁴)/ (2 × 2 × 2)

= (5t⁴)/ 8

(iii) (3⁵ × 10⁵ × 25)/ (5⁷ × 6⁵)

Solution:-

10⁵ can be written as = (5 × 2)⁵

= 5⁵ × 2⁵

25 can be written as = 5 × 5

= 5²

6⁵ can be written as = (2 × 3)⁵

= 2⁵ × 3⁵

Then we have,

= (3⁵ × 5⁵ × 2⁵ × 5²)/ (5⁷ × 2⁵ × 3⁵)

= (3⁵ × 5^{5 + 2} × 2⁵)/ (5⁷ × 2⁵ × 3⁵) … [∵a^m × aⁿ = a^{m + n}]

= (3⁵ × 5⁷ × 2⁵)/ (5⁷ × 2⁵ × 3⁵)

= (3^{5 – 5} × 5^{7 – 7} × 2^{5 – 5})

= (3⁰ × 5⁰ × 2⁰) … [∵a^m ÷ aⁿ = a^{m – n}]

= 1 × 1 × 1

= 1