Chapter 11: Exponents and Powers

Solution:-

(c) 3³ = 9

3 × 3 × 3 = 27

Therefore, 27 > 9

Solution:-

(b) 2

2⁶ = 2 × 2 × 2 × 2 × 2 × 2

= 64

8² = 8 × 8 = 64

In questions 23 to 39, fill in the blanks to make the statements true.

Solution:-

(-2)³¹ × (-2)¹³ = (-2)⁴⁴

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m × aⁿ = a^m+n

(-2)³¹ × (-2)¹³ = (-2)^{31 + 13}

= -2⁴⁴

Solution:-

(-3)⁸ ÷ (-3)⁵ = (-3)³

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m ÷ aⁿ = a^m–n , m>n

(-3)⁸ ÷ (-3)⁵ = (-3)^8-5

= -3³

Solution:-

(11/15)⁴ × (11/15)⁵ = (11/15)⁹

We know that,

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m × aⁿ = a^m+n

Therefore, (11/15)⁴ × (11/15)⁵ = (11/15)⁹

Solution:-

(-1/4)³ × (-1/4)⁸ = (-1/4)¹¹

We know that,

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m × aⁿ = a^m+n

Let us assume missing power be y,

Then, (-1/4)³ × (-1/4)^y = (-1/4)¹¹

(-1/4)^y = (-1/4)¹¹/(-1/4)³

(-1/4)^y = (-1/4)^{11 – 3}

(-1/4)^y = (-1/4)⁸

Solution:-

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

(a^m)ⁿ = a^mn

= (7/11)¹²

Solution:-

(6/13)¹⁰ ÷ [(6/13)⁵]² = (6/13)⁰

In the given question first consider [(6/13)⁵]²

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

(a^m)ⁿ = a^mn

= (6/13)¹⁰

Then, (6/13)¹⁰ ÷ (6/13)¹⁰

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m ÷ aⁿ = a^m–n , m>n

(6/13)¹⁰ ÷ (6/13)¹⁰ = (6/13)^{10 – 10}

= (6/13)⁰

Solution:-

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

(a^m)ⁿ = a^mn

= (-1/4)³²

Solution:-

(13/14)⁵ ÷ (13/14)² = (13/14)³

We know that,

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m ÷ aⁿ = a^m–n , m>n

Therefore, (13/14)⁵ ÷ (13/14)²= (13/14)^{5 – 2}

= (13/14)³