RD Sharma Solutions for Class 7 Maths Chapter 6 Exponents
By Ankit Gupta
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Updated on 15 Apr 2025, 12:35 IST
Exponents, also known as powers, are an essential concept in mathematics that help simplify the process of multiplying large numbers. In Class 7, Chapter 6 of RD Sharma Maths, students are introduced to the basics of exponents, which play a significant role in understanding higher-level mathematics.
Exponents provide a shorthand method for representing repeated multiplication. For example, instead of writing 2 × 2 × 2, we can simply write it as 2³. The number "2" is called the base, and "3" is the exponent, indicating that 2 is multiplied by itself three times. Understanding exponents makes it easier to work with large numbers and perform operations involving powers in algebra, geometry, and other branches of mathematics.
In this chapter, students will learn about the laws of exponents, which are the rules that help in simplifying expressions involving powers. These laws include important rules like the product of powers rule, the quotient of powers rule, and the power of a power rule. For instance, when multiplying two numbers with the same base, we simply add their exponents. This helps in quickly solving mathematical problems that involve powers.
This chapter also covers how to simplify expressions with exponents and how to deal with both positive and negative exponents. It helps students understand that exponents don’t just apply to positive numbers; they also work with fractions and negative numbers. For example, a negative exponent represents the reciprocal of the base raised to the positive exponent.
RD Sharma's solutions for Class 7 Maths Chapter 6 Exponents provide step-by-step explanations, ensuring that students grasp these concepts clearly. The solutions break down complex problems into manageable steps, making it easier for students to understand and solve problems involving exponents. By practicing the exercises in this chapter, students can build a solid foundation for more advanced topics in algebra and arithmetic.
In conclusion, mastering exponents is crucial as they form the building blocks for solving higher-level mathematical problems. This chapter offers a comprehensive introduction to the world of exponents, and with the help of RD Sharma solutions, students can confidently tackle problems involving powers and exponents. Whether it’s simplifying expressions or solving real-life problems, understanding exponents will help students become more confident and skilled in mathematics.
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Access Answers to RD Sharma Solutions for Class 7 Maths Chapter 6 Exponents
In this chapter, students will learn about decimals and how to perform basic operations with them. The solutions provided here are detailed and easy to follow, helping students understand each concept thoroughly.
1. Simplify using exponent laws and express the result in exponential form:
(i) 23 × 24 × 25
By the first exponent law, we know that:
am × an = am+n
So, 23 × 24 × 25 = 2(3+4+5) = 212.
(ii) 512 ÷ 53
Using the rule for division of exponents:
am ÷ an = am-n
512 ÷ 53 = 512 - 3 = 59.
(iii) (72)3
Using the rule for exponents inside parentheses:
(am)n = amn
(72)3 = 76.
(iv) (32)5 ÷ 34
Using the exponent rules, we get:
(32)5 ÷ 34 = 310 ÷ 34 = 3(10-4) = 36.
(v) 37 × 27
Using the rule that am × bm = (a × b)m:
37 × 27 = (3 × 2)7 = 67.
(vi) (521 ÷ 513) × 57
Using the exponent division rule:
521 ÷ 513 = 521-13 = 58.
Then, 58 × 57 = 5(8+7) = 515.
2. Simplify and express in exponential form:
(i) {(23)4 × 28} ÷ 212
{(23)4 × 28} ÷ 212 = 212 × 28 ÷ 212 = 220 ÷ 212 = 2(20-12) = 28.
(ii) (82 × 84) ÷ 83
(82 × 84) ÷ 83 = 8(2+4) ÷ 83 = 86 ÷ 83 = 8(6-3) = 83 = 29.
(iii) (57/52) × 53
5(7-2) × 53 = 55 × 53 = 5(5+3) = 58.
(iv) (54 × x10y5) ÷ (54 × x7y4)
(54-4 × x10-7y5-4) = x3y1 = 1 × x3y.
3. Simplify and express in exponential form:
(i) {(32)3 × 26} × 56
{(32)3 × 26} × 56 = 36 × 26 × 56 = 306.
(ii) (x/y)12 × y24 × (23)4
(x/y)12 × y24-12 × 212 = x12 × y12 × 212 = (2xy)12.
(iii) (5/2)6 × (5/2)2
(5/2)6+2 = (5/2)8.
(iv) (2/3)5 × (3/5)5
(2/3)5 × (3/5)5 = (2/5)5.
4. Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3.
9 × 9 × 9 × 9 × 9 = (9)5 = (32)5 = 310.
5. Simplify and express in exponential form:
(i) (25)3 ÷ 53
(52)3 ÷ 53 = 56 ÷ 53 = 53.
(ii) (81)5 ÷ (32)5
(81)5 ÷ (32)5 = (34)5 ÷ 310 = 320 ÷ 310 = 310.
(iii) 98 × (x2)5 ÷ (27)4 × (x3)2
316 × x10 ÷ 312 × x6 = 34 × x4 = (3x)4.
(iv) 32 × 78 × 136 ÷ 212 × 913
32 × 7276 × 136 ÷ 212 × 133 × 73 = 916 ÷ 913 = 913.
6. Simplify:
(i) (35)11 × (315)4 – (35)18 × (35)5
3115 – 3115 = 0.
(ii) (16 × 2n+1 – 4 × 2n) ÷ (16 × 2n+2 – 2 × 2n+2)
2n(23 – 1) / 2n(24 – 1) = (1/2).
(iii) (10 × 5n+1 + 25 × 5n) ÷ (3 × 5n+2 + 10 × 5n+1)
15/25 = 3/5.
(iv) (16)7 × (25)5 × (81)3 ÷ (15)7 × (24)5 × (80)3
24/8 = 2.
7. Find the value of n in each of the following:
(i) 52n × 53 = 511
n = 4.
(ii) 9 × 3n = 37
n = 5.
(iii) 8 × 2n+2 = 32
n = 0.
(iv) 72n+1 ÷ 49 = 73
n = 2.
(v) (3/2)4 × (3/2)5 = (3/2)2n+1
n = 4.
(vi) (2/3)10 × {(3/2)2}5 = (2/3)2n-2
n = 1.
FAQs on RD Sharma Solutions for Class 7 Maths Chapter 6 Exponents
What are exponents in mathematics?
Exponents, also known as powers, are a way to represent repeated multiplication of the same number. For example, 2³ means 2 × 2 × 2, where 2 is the base and 3 is the exponent.
Why are exponents important in mathematics?
Exponents help simplify complex expressions and make it easier to work with large numbers. They are essential in algebra, geometry, and various other branches of mathematics.
What is the product of powers rule?
The product of powers rule states that when multiplying two numbers with the same base, you add their exponents. For example, a³ × a² = a(3+2) = a⁵.
How do negative exponents work?
A negative exponent represents the reciprocal of the base raised to the positive exponent. For example, a⁻³ = 1/a³. Negative exponents allow us to work with fractions and simplify expressions.
How can I simplify expressions with exponents?
To simplify expressions with exponents, use the laws of exponents. These include rules for multiplying powers with the same base, dividing powers with the same base, and raising a power to another power.
How can RD Sharma solutions help in understanding exponents?
RD Sharma solutions offer step-by-step explanations for each problem, making it easier for students to understand and apply exponent laws. The solutions break down complex problems into simple, manageable steps.