Exponents is in general referred to as ‘powers’. It is a way to express repeated multiplication of the same number. This concept is important as it helps us simplify and manage large numbers with ease.
Exploring the Laws of Exponents
Multiplying Powers with the Same Base
When we multiply numbers with the same base, we add their exponents. This law turns complex multiplications into simple additions of exponents, making calculations quicker.
Example: 2⁴ × 2³
2^(4+3) = 2⁷
When multiplying powers with the same base, add their exponents.
Dividing Powers with the Same Base
Similar to multiplication, when we divide numbers with the same base, we subtract the exponents. This law is helpful in breaking down complex divisions into manageable pieces.
Example: 5⁶ ÷ 5²
5^(6-2) = 5⁴
When dividing powers with the same base, subtract the exponent of the denominator from the exponent of the numerator.
Taking Power of a Power
This involves multiplying exponents when a number raised to a power is further raised to another power. It’s a concept that shows how powers can be compounded.
Example: (3²)⁴
3^(2×4) = 3⁸
When taking a power of a power, multiply the exponents.
Multiplying and Dividing Powers with the Same Exponents
When we multiply or divide powers with the same exponents, we apply the exponent to the result of the multiplication or division of the bases.
Example(Multiplying): 4³ × 6³
(4×6)³ = 24³
When multiplying powers with the same exponent, multiply the bases and keep the exponent.
Example(Dividing): 10⁵ ÷ 2⁵
(10÷2)⁵ = 5⁵
When dividing powers with the same exponent, divide the bases and keep the exponent.
Numbers with Exponent Zero
Any number raised to the power of zero is 1. This intriguing concept is fundamental and often surprising at first.
Example: 7⁰
1
Any non-zero number raised to the power of zero equals 1.
Question and Answers for Class 7 Maths Exercise 11.2 Chapter 11 Exponents and Powers
Question 1. Using laws of exponents, simplify and write the answer in e×ponential form:
(i) 3² × 3⁴ × 3⁸
(ii) 6¹⁵ ÷ 6¹⁰
(iii) a³ × a²
(iv) 7ˣ × 7²
(v) (5²)³ ÷ 5³
(vi) 2⁵ × 5⁵
(vii) a⁴ × b⁴
(viii) (3⁴)³
(ix) (2²⁰ ÷ 2¹⁵) × 2³
(x) 8ᵗ ÷ 8²
Solution
(i) 3² × 3⁴ × 3⁸
= 3^(2+4+8) [Adding exponents for the same base]
= 3¹⁴
(ii) 6¹⁵ ÷ 6¹⁰
= 6^(15-10) [Subtracting exponents for the same base]
= 6⁵
(iii) a³ × a²
= a^(3+2) [Adding exponents for the same base]
= a⁵
(iv) 7ˣ × 7²
= 7^(x+2) [Adding exponents for the same base]
(v) (5²)³ ÷ 5³
= 5^(2×3) ÷ 5³ [Applying power of a power rule]
= 5⁶ ÷ 5³
= 5^(6-3) [Subtracting exponents for the same base]
= 5³
(vi) 2⁵ × 5⁵
= (2×5)⁵ [Multiplying before raising to the power]
= 10⁵
(vii) a⁴ × b⁴
= (a × b)⁴ [Multiplying before raising to the power]
(viii) (3⁴)³
= 3^(4×3) [Applying power of a power rule]
= 3¹²
(ix) (2²⁰ ÷ 2¹⁵) × 2³
= 2^(20-15) × 2³ [Subtracting exponents for the same base]
= 2⁵ × 2³
= 2⁸ [Adding exponents for the same base]
(x) 8ᵗ ÷ 8²
= 8^(t-2) [Subtracting exponents for the same base]
Question 2. Simplify and express each of the following in exponential form:
(i) (2³ × 3⁴ × 4) ÷ (3 × 32)
(ii) ((5²)³ × 5⁴) ÷ 5
(iii) 25⁴ ÷ 5³
(iv) (3 × 7² × 11⁸) (21 × 11³)
(v) 3⁷/(3⁴ × 3³)
(vi) 2⁰ + 3⁰ + 4⁰
(vii) 2⁰ × 3⁰ × 4⁰
(viii) (3⁰ + 2⁰) × 5⁰
(ix) (2⁸ × a⁵)/(4³ × a³)
(x) (a⁵/a³) × a⁸
(xi) (4⁵ × a⁸b³)/(4⁵ × a⁵b²)
(xii) (2³ × 2)²
Solution
(i) (2³ × 3⁴ × 4) ÷ (3 × 32)
= (2³ × 3⁴ × 2²) ÷ (3 × 3²) [Since 4 = 2²]
= (2³ × 2² × 3⁴) ÷ 3³ [Since 32 = 3²]
= 2⁵ × 3¹ [Combining the powers of 2 and reducing the powers of 3]
(ii) ((5²)³ × 5⁴) ÷ 5
= 5⁶ × 5⁴ ÷ 5 [Since (5²)³ = 5^(2×3) = 5⁶]
= 5¹⁰ ÷ 5 [Adding the powers of 5]
= 5⁹ [Reducing the power by 1]
(iii) 25⁴ ÷ 5³
= (5²)⁴ ÷ 5³ [Since 25 = 5²]
= 5⁸ ÷ 5³ [Since (5²)⁴ = 5^(2×4) = 5⁸]
= 5⁵ [Reducing the power by 3]
(iv) (3 × 7² × 11⁸) × (21 × 11³)
= 3 × 7² × 11⁸ × 3 × 7 × 11³ [Expanding 21 as 3 × 7]
= 3² × 7³ × 11¹¹ [Combining like bases and adding their exponents]
(v) 3⁷ ÷ (3⁴ × 3³)
= 3⁷ ÷ 3⁷ [Since 3⁴ × 3³ = 3^(4+3) = 3⁷]
= 3⁰ [Dividing like bases with same exponents]
= 1 [Any number to the power of 0 is 1]
(vi) 2⁰ + 3⁰ + 4⁰
= 1 + 1 + 1 [Since any number to the power of 0 is 1]
= 3
(vii) 2⁰ × 3⁰ × 4⁰
= 1 × 1 × 1 [Since any number to the power of 0 is 1]
= 1
(viii) (3⁰ + 2⁰) × 5⁰
= (1 + 1) × 1 [Since any number to the power of 0 is 1]
= 2 × 1
= 2
(ix) (2⁸ × a⁵) ÷ (4³ × a³)
= 2⁸ × a⁵ ÷ (2⁶ × a³) [Since 4³ = (2²)³ = 2⁶]
= 2² × a² [Reducing the powers by division]
(x) (a⁵ ÷ a³) × a⁸
= a² × a⁸ [Since a⁵ ÷ a³ = a^(5-3) = a²]
= a¹⁰ [Adding the exponents]
(xi) (4⁵ × a⁸b³) ÷ (4⁵ × a⁵b²)
= a⁸b³ ÷ a⁵b² [4⁵ cancels out]
= a³b [Reducing the powers by division]
(xii) (2³ × 2)²
= 2⁴² [Since 2³ × 2 = 2^(3+1) = 2⁴]
= 2⁸ [Applying the power of a power rule]
Question 3. Say true or false and justify your answer:
(i) 10 × 10¹¹ = 100¹¹
(ii) 2³ > 5²
(iii) 2³ × 3² = 6⁵
(iv) 32⁰ = (1000)⁰
Solution
(i) 10 × 10¹¹ = 100¹¹
False.
10 × 10¹¹ = 10^(1+11) = 10¹².
100¹¹ is 100 raised to the power 11, which is not the same as 10¹².
(ii) 2³ > 5²
False.
2³ = 2 × 2 × 2 = 8.
5² = 5 × 5 = 25.
8 is not greater than 25.
(iii) 2³ × 3² = 6⁵
False.
2³ × 3² = 8 × 9 = 72.
6⁵ = 6 × 6 × 6 × 6 × 6 = 7776.
72 is not equal to 7776.
(iv) 32⁰ = (1000)⁰
True.
Any number raised to the power of 0 is 1.
So, both 32⁰ and (1000)⁰ equal 1.
Question 4. Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
(ii) 270
(iii) 729 × 64
(iv) 768
Solution
(i) 108 × 192
108 = 2² × 3³ [Prime factorization of 108]
192 = 2⁶ × 3 [Prime factorization of 192]
108 × 192 = 2² × 3³ × 2⁶ × 3
= 2⁸ × 3⁴
(ii) 270
270 = 2 × 3³ × 5 [Prime factorization of 270]
(iii) 729 × 64
729 = 3⁶ [Prime factorization of 729]
64 = 2⁶ [Prime factorization of 64]
729 × 64 = 3⁶ × 2⁶
(iv) 768
768 = 2⁸ × 3 [Prime factorization of 768]
Question 5. Simplify:
(i) ((2⁵)² × 7³)/(8³ × 3)
(ii) (25 × 5² × t⁸)/(10³ × t⁴)
(iii) (3⁵ × 10⁵ × 25)/(5⁷ × 6⁵)
Solution
(i) ((2⁵)² × 7³) ÷ (8³ × 3)
= (2¹⁰ × 7³) ÷ (2³³ × 3) [Since 8³ = (2³)³ = 2⁹]
= 2¹⁰ × 7³ ÷ 2⁹ ÷ 3
= 2¹ × 7³ ÷ 3
(ii) (25 × 5² × t⁸) ÷ (10³ × t⁴)
= (5² × 5² × t⁸) ÷ (2³ × 5³ × t⁴) [Since 25 = 5² and 10³ = 2³ × 5³]
= 5⁴ × t⁸ ÷ 2³ × 5³ × t⁴
= 5¹ × t⁴ ÷ 2³
(iii) (3⁵ × 10⁵ × 25) ÷ (5⁷ × 6⁵)
= (3⁵ × 2⁵ × 5⁵ × 5²) ÷ (5⁷ × 2⁵ × 3⁵) [Since 10⁵ = 2⁵ × 5⁵ and 25 = 5²]
= 5⁷ × 2⁵ × 3⁵ ÷ 5⁷ × 2⁵ × 3⁵
= 1 [All terms cancel out]
