12.- Exponents and Powers

Exercise 12.1

Question 1- Evaluate

(i) 3⁻² (ii) (−4)⁻² (iii)

Answer - (i)

(ii)

(iii)

Question 2- Simplify and express the result in power notation with positive exponent.

(i) (ii)

(iii) (iv)

(v)

Answer - (i) (−4)⁵ ÷ (−4)⁸ = (−4)^{5 − 8} (a^m ÷ aⁿ = a^m − n)

= (− 4)⁻³

(ii)

(iii)

(iv) (3^{− 7} ÷ 3⁻¹⁰) × 3⁻⁵ = (3^{−7 − (−10)}) × 3⁻⁵ (a^m ÷ aⁿ = a^m − n)

= 3³ × 3⁻⁵

= 3^{3 + (− 5)} (a^m × aⁿ = a^m + n)

= 3⁻²

(v) 2⁻³ × (−7)⁻³ = 

Question 3 - Find the value of.

(i) (3⁰ + 4⁻¹) × 2² (ii) (2⁻¹ × 4⁻¹) ÷2⁻²

(iii)  (iv) (3⁻¹ + 4⁻¹ + 5⁻¹)⁰

(v)

Answer -  (i)

(ii) (2⁻¹ × 4⁻¹) ÷ 2^{− 2} = [2⁻¹ × {(2)²}^{− 1}] ÷ 2^{− 2}

= (2^{− 1} × 2^{− 2}) ÷ 2^{− 2} 

= 2^{−1+ (−2)} ÷ 2⁻² (a^m × aⁿ = a^m + n)

= 2⁻³ ÷ 2⁻²

= 2^{−3 − (−2)} (a^m ÷ aⁿ = a^m − n)

= 2^{−3 + 2} = 2 ⁻¹

(iii)

(iv) (3⁻¹ + 4⁻¹ + 5⁻¹)⁰ 

= 1 (a⁰ = 1)

(v)

Question 4- Evaluate (i) (ii)

Answer - (i)

(ii)

Question 5- Find the value of m for which 5^m ÷5⁻³ = 5⁵.

Answer -  5^m ÷ 5⁻³ = 5⁵

5^{m − (− 3)} = 5⁵ (a^m ÷ aⁿ = a^m − n)

5^{m + 3} = 5⁵

Since the powers have same bases on both sides, their respective exponents must be equal.

m + 3 = 5

m = 5 − 3

m = 2

Question 6- Evaluate (i) (ii)

Answer - (i)

(ii)

Question 7- Simplify. (i) (ii)

Answer - (i)

(ii)

Exercise -12.2

Question 1- Express the following numbers in standard form.

(i) 0.0000000000085 (ii) 0.00000000000942

(iii) 6020000000000000 (iv) 0.00000000837

(v) 31860000000

Answer - (i) 0.0000000000085 = 8.5 × 10⁻¹²

(ii) 0.00000000000942 = 9.42 × 10⁻¹²

(iii) 6020000000000000 = 6.02 × 10¹⁵

(iv) 0.00000000837 = 8.37 × 10⁻⁹

(v) 31860000000 = 3.186 × 10¹⁰

^{Question 2 -} Express the following numbers in usual form.

(i) 3.02 × 10⁻⁶ (ii) 4.5 × 10⁴

(iii) 3 × 10⁻⁸ (iv) 1.0001 × 10⁹

(v) 5.8 × 10¹² (vi) 3.61492 × 10⁶

^{Answer -} (i) 3.02 × 10⁻⁶ = 0.00000302

(ii) 4.5 × 10⁴ = 45000

(iii) 3 × 10⁻⁸ = 0.00000003

(iv) 1.0001 × 10⁹ = 1000100000

(v) 5.8 × 10¹² = 5800000000000

(vi) 3.61492 × 10⁶ = 3614920

Question 3- Express the number appearing in the following statements in standard form.

(i) 1 micron is equal to m.

(ii) Charge of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.

(iii) Size of a bacteria is 0.0000005 m

(iv) Size of a plant cell is 0.00001275 m

(v) Thickness of a thick paper is 0.07 mm

Answer - (i)  = 1 × 10⁻⁶

(ii) 0.000, 000, 000, 000, 000, 000, 16 = 1.6 × 10⁻¹⁹

(iii) 0.0000005 = 5 × 10⁻⁷

(iv) 0.00001275 = 1.275 × 10⁻⁵

(v) 0.07 = 7 × 10⁻²

^{Question 4-}  In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?

Answer -  Thickness of each book = 20 mm

Hence, thickness of 5 books = (5 × 20) mm = 100 mm

Thickness of each paper sheet = 0.016 mm

Hence, thickness of 5 paper sheets = (5 × 0.016) mm = 0.080 mm

Total thickness of the stack = Thickness of 5 books + Thickness of 5 paper sheets

= (100 + 0.080) mm

= 100.08 mm

= 1.0008 × 10² mm