Chapter 11: Exponents and Powers

Solution:-

(b) (-3)⁶

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

(a^m)ⁿ = a^mn

= -3⁶

Solution:-

(b) x⁶

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

a^m ÷ aⁿ = a^m–n , m>n

so, x⁸ ÷ x²

= x⁸/x²

= x^{8 – 2}

= x⁶

Solution:-

(c) fifth power of x

From the question, it is given that,

The square of x = x²

The cube of x = x³

As per the condition given in the question, x² × x³

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

a^m × aⁿ = a^m+n

x² × x³ = x²⁺³

= x⁵

Therefore, fifth power of x.

Solution:-

(c) (x÷y)⁵

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

a^m ÷ b^m = (a/b)^m

x⁵ ÷ y⁵ = (x/y)⁵ = (x ÷ y)⁵

Solution:-

(c) a^m+n

a^m × aⁿ is equal to a^{m + n}

Solution:-

(c) 3

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

a⁰ = 1

So, 1⁰ = 1, 2⁰ = 1, 3^o = 1

Then, (1 + 1 + 1)

= 3

Solution:-

(c) 101

By splitting, (10²² + 10²⁰)/10²⁰ we get,

= (10²²/10²⁰) + (10²⁰/10²⁰)

By using the rule a^m ÷ aⁿ = a^m–n, m>n

= 10^{(22 – 20)} + 1

= 10² + 1

= 100 + 1

= 101

Solution:-

(d) 1.2345 × 10⁴

Any number can be expressed as a decimal number between 1.0 and 10.0 (including 1.0) multiplied by a power of 10. Such form of a number is called its standard form or scientific notation.

Solution:-

(c) 3

2¹⁹⁹⁸ – 2¹⁹⁹⁷ – 2¹⁹⁹⁶ + 2¹⁹⁹⁵ = K2¹⁹⁹⁵

By cross multiplication we get,

(2¹⁹⁹⁸/2¹⁹⁹⁵) – (2¹⁹⁹⁷/2¹⁹⁹⁵) – (2¹⁹⁹⁶/2¹⁹⁹⁵) + (2¹⁹⁹⁵/2¹⁹⁹⁵) = K

By the rule, a^m ÷ aⁿ = a^m–n , m>n

K = (2^{1998 – 1995}) – (2^{1997 – 1995}) – (2^{1996 – 1995}) + (2^{1995 – 1995})

K = 2³ – 2² – 2¹ + 2⁰

K = 8 – 4 – 2 + 1

K = 3

Solution:-

(b) 2⁰ × 3⁰ × 4⁰

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

a⁰ = 1

So, 2⁰ = 1, 3⁰ = 1, 4^o = 1

Then, (1 × 1 × 1)

= 1