Class 8 Maths - Exponents and Powers NCERT Solutions

Chapter 12: Exponents and Powers (NCERT Solutions)

Exercise 12.1

Q1. Evaluate:
(i) 3^-2 (ii) (-4)^-2 (iii) (1/2)^-5

(i) 3^-2
Using a^-m = 1/a^m
3^-2 = 1 / 3² = 1 / 9

(ii) (-4)^-2
(-4)^-2 = 1 / (-4)² = 1 / 16 = 1 / 16

(iii) (1/2)^-5
(1/2)^-5 = (2/1)⁵ = 2⁵ = 32

Q2. Simplify and express the result in power notation with positive exponent:
(i) (-4)⁵ ÷ (-4)⁸
(ii) (1 / 2³)²
(iii) (-3)⁴ × (5/3)⁴
(iv) (3^-7 ÷ 3^-10) × 3^-5
(v) 2^-3 × (-7)^-3

(i) (-4)⁵ ÷ (-4)⁸ = (-4)^5-8 = (-4)^-3
Positive exponent form: 1 / (-4)³.

(ii) (1 / 2³)² = 1² / (2³)² = 1 / 2⁶.

(iii) (-3)⁴ × (5/3)⁴ = (-1 × 3)⁴ × (5⁴ / 3⁴)
= (-1)⁴ × 3⁴ × (5⁴ / 3⁴) = 1 × 5⁴ = 5⁴.

(iv) (3^-7 ÷ 3^-10) × 3^-5 = (3^{-7 - (-10)}) × 3^-5 = (3³) × 3^-5
= 3^{3 + (-5)} = 3^-2 = 1 / 3².

(v) 2^-3 × (-7)^-3 = [2 × (-7)]^-3 = (-14)^-3
Positive exponent form: 1 / (-14)³.

Q3. Find the value of:
(i) (3⁰ + 4^-1) × 2²
(ii) (2^-1 × 4^-1) ÷ 2^-2
(iii) (1/2)^-2 + (1/3)^-2 + (1/4)^-2
(iv) (3^-1 + 4^-1 + 5^-1)⁰
(v) {(-2/3)^-2}²

(i) (1 + 1/4) × 4 = (5/4) × 4 = 5.

(ii) (1/2 × 1/4) ÷ (1/2²) = (1/8) ÷ (1/4) = 1/8 × 4 = 1/2.

(iii) (2)² + (3)² + (4)² = 4 + 9 + 16 = 29.

(iv) Any non-zero number raised to the power 0 is 1. Thus, the value is 1.

(v) {(-2/3)^-2}² = (-2/3)^-4 = (-3/2)⁴ = (-3)⁴ / (2)⁴ = 81 / 16.

Q4. Evaluate:
(i) (8^-1 × 5³) / 2^-4
(ii) (5^-1 × 2^-1) × 6^-1

(i) (8^-1 × 5³) / 2^-4 = (1/8 × 125) / (1/16)
= (125 / 8) × 16 = 125 × 2 = 250.
Alternatively: 8^-1 = (2³)^-1 = 2^-3. Expression = 2^-3 × 5³ × 2⁴ = 2¹ × 5³ = 2 × 125 = 250.

(ii) (1/5 × 1/2) × 1/6 = (1/10) × 1/6 = 1 / 60.

Q5. Find the value of m for which 5^m ÷ 5^-3 = 5⁵.

5^m ÷ 5^-3 = 5^{m - (-3)} = 5^{m + 3}.
Given, 5^{m + 3} = 5⁵.
Comparing both sides, m + 3 = 5 ⇒ m = 5 - 3 = 2.

Q6. Evaluate:
(i) { (1/3)^-1 - (1/4)^-1 }^-1
(ii) (5/8)^-7 × (8/5)^-4

(i) { 3 - 4 }^-1 = {-1}^-1 = 1 / (-1) = -1.

(ii) (5/8)^-7 × (8/5)^-4 = (8/5)⁷ × (8/5)^-4
= (8/5)^{7 + (-4)} = (8/5)³ = 8³ / 5³ = 512 / 125.

Q7. Simplify:
(i) (25 × t^-4) / (5^-3 × 10 × t^-8) (t ≠ 0)
(ii) (3^-5 × 10^-5 × 125) / (5^-7 × 6^-5)

(i) (5² × t^-4) / (5^-3 × (5 × 2) × t^-8)
= (5² × t^-4) / (5^-2 × 2 × t^-8)
= (5^{2 - (-2)} × t^{-4 - (-8)}) / 2
= (5⁴ × t⁴) / 2 = 625t⁴ / 2.

(ii) (3^-5 × (2 × 5)^-5 × 5³) / (5^-7 × (2 × 3)^-5)
= (3^-5 × 2^-5 × 5^-5 × 5³) / (5^-7 × 2^-5 × 3^-5)
= [3^-5 / 3^-5] × [2^-5 / 2^-5] × [5^-5+3 / 5^-7]
= 1 × 1 × 5^{-2 - (-7)}
= 5⁵ = 3125.

Exercise 12.2

Q1. Express the following numbers in standard form.
(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020000000000000
(iv) 0.00000000837
(v) 31860000000

(i) Move decimal 12 places to the right: 8.5 × 10^-12.

(ii) Move decimal 12 places to the right: 9.42 × 10^-12.

(iii) Move decimal 15 places to the left: 6.02 × 10¹⁵.

(iv) Move decimal 9 places to the right: 8.37 × 10^-9.

(v) Move decimal 10 places to the left: 3.186 × 10¹⁰.

Q2. 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

(i) Move decimal 6 places to the left: 0.00000302.

(ii) Move decimal 4 places to the right: 45000.

(iii) Move decimal 8 places to the left: 0.00000003.

(iv) Move decimal 9 places to the right: 1000100000.

(v) Move decimal 12 places to the right: 5800000000000.

(vi) Move decimal 6 places to the right: 3614920.

Q3. Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to 1/1000000 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.

(i) 1/1000000 = 1 / 10⁶ = 1 × 10^-6 m.

(ii) Move decimal 19 places right: 1.6 × 10^-19 coulomb.

(iii) Move decimal 7 places right: 5.0 × 10^-7 m.

(iv) Move decimal 5 places right: 1.275 × 10^-5 m.

(v) Move decimal 2 places right: 7 × 10^-2 mm.

Q4. 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?

Thickness of 5 books = 5 × 20 mm = 100 mm.
Thickness of 5 paper sheets = 5 × 0.016 mm = 0.080 mm.
Total thickness = 100 mm + 0.080 mm = 100.08 mm.
In standard form: 1.0008 × 10² mm.

Class 8 Maths - Exponents and Powers Practice Questions

Chapter 12: Exponents and Powers (Practice Questions)

RD Sharma / Extra Practice Questions

Q1. Evaluate: (1/2)^-5

(1/2)^-5 = (2/1)⁵ = 2⁵
= 2 × 2 × 2 × 2 × 2 = 32.

Q2. Simplify and express the result in power notation with positive exponent: (-4)⁵ ÷ (-4)⁸

Using a^m ÷ aⁿ = a^m-n
(-4)⁵ ÷ (-4)⁸ = (-4)^{5 - 8} = (-4)^-3
To make the exponent positive: 1 / (-4)³.

Q3. Find the value of m for which 5^m ÷ 5^-3 = 5⁵

5^m ÷ 5^-3 = 5⁵
5^{m - (-3)} = 5⁵
5^{m + 3} = 5⁵
On comparing the exponents on both sides (since bases are equal):
m + 3 = 5
m = 5 - 3 = 2.

Q4. Express 0.00000000837 in standard form.

Move the decimal point 9 places to the right.
So, exponent of 10 will be -9.
Standard form = 8.37 × 10^-9.

Q5. Simplify: [(1/3)^-2 - (1/2)^-3] ÷ (1/4)^-2

[(1/3)^-2 - (1/2)^-3] = [3² - 2³]
= [9 - 8] = 1.
Now divide by (1/4)^-2 = 4² = 16.
Result = 1 ÷ 16 = 1/16.

Q6. Write 3.02 × 10^-6 in usual form.

3.02 × 10^-6 = 3.02 / 1000000
Move the decimal point 6 places to the left.
Result = 0.00000302.

Q7. By what number should (-4)^-1 be multiplied so that the product becomes 10^-1?

Let the required number be x.
(-4)^-1 × x = 10^-1
(-1/4) × x = 1/10
x = (1/10) / (-1/4) = (1/10) × (-4/1) = -4/10 = -2/5.

Q8. Find the value of x if (2/5)³ × (2/5)^-6 = (2/5)^{2x - 1}

(2/5)^{3 + (-6)} = (2/5)^{2x - 1}
(2/5)^-3 = (2/5)^{2x - 1}
Equating exponents: -3 = 2x - 1
2x = -3 + 1 = -2
x = -2 / 2 = -1.

Q9. Simplify: (3^-5 × 10^-5 × 125) / (5^-7 × 6^-5)

Convert everything to prime bases:
10^-5 = (2 × 5)^-5 = 2^-5 × 5^-5
125 = 5³
6^-5 = (2 × 3)^-5 = 2^-5 × 3^-5
Expression = (3^-5 × 2^-5 × 5^-5 × 5³) / (5^-7 × 2^-5 × 3^-5)
= [3^-5 / 3^-5] × [2^-5 / 2^-5] × [5^-5+3 / 5^-7]
= 1 × 1 × 5^{-2 - (-7)}
= 5⁵ = 3125.

Q10. Evaluate: (5^-1 × 2^-1) ÷ 6^-1

(1/5 × 1/2) ÷ (1/6)
= (1/10) × (6/1) = 6/10 = 3/5.

Q11. The size of a plant cell is 0.00001275 m. Express it in standard form.

Move the decimal point 5 places to the right.
1.275 × 10^-5 m.

Q12. If a = 2 and b = 3, find the value of (a^b + b^a)^-1.

a^b + b^a = 2³ + 3² = 8 + 9 = 17.
(17)^-1 = 1/17.

Q13. Express 4.5 × 10⁴ in usual form.

4.5 × 10000 = 45000.

Q14. Simplify: (t^-4) / (t^-8)

t^{-4 - (-8)} = t^{-4 + 8}
= t⁴.

Q15. Write the multiplicative inverse of 10^-100.

The multiplicative inverse of a^m is a^-m.
For 10^-100, it is 1 / 10^-100 = 10¹⁰⁰.

Q16. Find the value of [ (1/4)^-1 - (1/3)^-1 ]^-1

(1/4)^-1 = 4
(1/3)^-1 = 3
[4 - 3]^-1 = [1]^-1 = 1.

Q17. Simplify: (2^-1 × 4^-1) ÷ 2^-2

4^-1 = (2²)^-1 = 2^-2.
Numerator = 2^-1 × 2^-2 = 2^-3.
Division = 2^-3 ÷ 2^-2 = 2^{-3 - (-2)} = 2^-1 = 1/2.

Q18. Express 65300000 in standard form.

Move the decimal point 7 places to the left.
6.53 × 10⁷.

Q19. Find x if 8^{x - 1} = 2^{x + 3}.

Write 8 as 2³.
(2³)^{x - 1} = 2^{x + 3}
2^{3x - 3} = 2^{x + 3}
Equating powers: 3x - 3 = x + 3
2x = 6 ⇒ x = 3.

Q20. Evaluate: ( -3/4 )^-3

( -3/4 )^-3 = ( -4/3 )³
= (-4)³ / (3)³ = -64 / 27.

Class 8 Maths - Exponents and Powers Summary

Chapter 12: Exponents and Powers (Concepts & Formulas)

1. Introduction to Exponents

An expression that represents repeated multiplication of the same factor is called a power.
For example, in 10⁴, 10 is the base and 4 is the exponent. It means 10 × 10 × 10 × 10.

Numbers with negative exponents: A negative exponent means reciprocal of the base raised to the positive exponent.
a^-m = 1 / a^m (where a is any non-zero integer).

a^-m is the multiplicative inverse of a^m.

2. Laws of Exponents

If a and b are non-zero integers and m, n are integers, then:

Multiplication (same base): a^m × aⁿ = a^m+n
Division (same base): a^m ÷ aⁿ = a^m-n
Power of a power: (a^m)ⁿ = a^mn
Multiplication (same exponent): a^m × b^m = (ab)^m
Division (same exponent): a^m ÷ b^m = (a/b)^m
Zero Exponent: a⁰ = 1

3. Standard Form of Numbers

Very large and very small numbers can be expressed in standard form (scientific notation).

A number is in standard form when it is written as k × 10ⁿ, where 1 ≤ k < 10 and n is an integer.

Very large numbers: The decimal point is moved to the left, and the exponent of 10 is positive (equal to the number of places the decimal point moved).
E.g., 150,000,000,000 = 1.5 × 10¹¹
Very small numbers: The decimal point is moved to the right, and the exponent of 10 is negative.
E.g., 0.000007 = 7.0 × 10^-6

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