In Class 7 Maths Chapter 11, Exponents and Powers, we will learn about the concept of exponents. In daily life or in scientific calculations, whenever we come across large numbers which are not feasible to be noted in their basic units, then, exponents come into use. To make these numbers easy to read, understand, and compare, we use exponents. Additionally, we will learn the Laws of the Exponents, Powers, and the Decimal Number System. Let us now have a look at the NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Notes and Solutions (PDF available).
Download NCERT Class 7 Maths Chapter 11 – Exponents and Powers Notes and Solutions PDF
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Download NCERT Class 7 Maths Chapter 11 – Exponents and Powers Notes and Solutions PDF
NCERT Solutions Class 7 Maths Chapter 11 – PDF Available
Check the topic-wise notes for NCERT Solutions Maths Class 7 Chapter 11, Exponents and Powers below. You can also download the PDF of the notes and take a printout to study later when you need quick revision before going to the e𝑥am hall.
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Topic 1: Exponents
The concept of exponents can be explained by using an example. The number 104 is read as 10 raised to the power of 4 or simply as the fourth power of 10. Here, 104 is called the exponential form of 10,000.
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Topic 2: Laws of Exponents
- Multiplication of Powers with the Same Base: Whenever the same bases with different or same powers are multiplied, their powers get added while the base number remains the same.
For example: 42 × 43 = 42+3 = 45
In general terms, this can be noted as: am × an = am + n
- Division of Powers with the Same Base: Whenever the same bases with different or same powers are divided, their powers get subtracted while the base number remains the same.
For example: 42 ÷ 43 = 42-3 = 4-1 = ¼
In general terms, this can be noted as: am ÷ an = am – n
- Multiplication of Powers with Different Base: For any non-zero integer “a”, am × bm = (ab)m.
- Division of Powers with Different Base: For any non-zero integer “a”, am ÷ bm = (a/b)m.
Topic 3: Taking Power of a Power
Whenever a power is raised to an exponent, then, both the powers are multiplied to give the final answer.
For example, in (43)4, the exponent 43 is raised to the power of 4. Hence, the value of this expression can be calculated as 43 × 4 = 412
This can be understood with the following general formula: (am)n = amn.
Download NCERT Class 7 Maths Chapter 11 – Exponents and Powers Notes and Solutions PDF
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NCERT Solutions of Class 7 Maths Chapter 11: Exponents and Powers- PDF Available!
Below we have provided solutions for NCERT Solutions Class 7 Maths Chapter 11, Exponents and Powers. Go through for answers to some important questions.
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E𝑥ercise 11.1 Solutions
Q 1. Find the value of 112.
Solutions. The value of 112 can be calculated as:
11 × 11 = 121
Hence, 112 = 121
Q 2. Express each of the following numbers using exponential notation:
- 343
- 512
Solutions. To write the given numbers in exponential forms, we need to find out their factors.
- The factors of 343 are 7 × 7 × 7
∴ the exponential form of 343 is 73.
- The factors of 512 are 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
∴ the exponential form of 512 is 29.
Q 3. Express each of the following as a product of the powers of their prime factors:
- 648
- 405
Solutions. The answers are given below.
- The factors of 648 are 2 × 2 × 2 × 3 × 3 × 3 × 3
∴ the desired form of 648 is 23 × 34.
- The factors of 405 are 3 × 3 × 3 × 3 × 5
∴ the desired form of 648 is 34 × 5.
Exercise 11.2 Solutions
Q 1. Using laws of exponents, simplify and write the answer in exponential form:
- 32 × 34 × 38
- 615 ÷ 610
- 7𝑥 × 72
- (34)3
Solutions. The solutions are given below.
- The base values in all the given exponents are the same, i.e. 3. We know that am × an = am + n
∴ 32 × 34 × 38 = 32 + 4 + 8 = 314
- The base values in all the given exponents are the same, i.e. 6. We know that am ÷ an = am – n
∴ 615 ÷ 610 = 615 – 10 = 65
- The base values in all the given exponents are the same, i.e. 7. We know that am × an = am + n
∴ 7𝑥 × 72 = 7(𝑥 + 2)
- From the general formula of taking the power of a power, we know that (am)n = amn.
∴ (34)3 = 34 × 3 = 312
Q 2. Simplify and express each of the following in exponential form:
- 20 × 30 × 40
- (23 × 2)2
- 254 ÷ 53
Solutions: The solutions are given below.
- We know that the value of any constant raised to the power of 0 is 1.
∴ 20 × 30 × 40 = 1 × 1 × 1 = 1
- We know that am × an = am + n
∴ (23 × 2)2 = (23 × 21)2
⇒ (23 + 1)2 = (24)2
Now, from taking the power of power, we know that (am)n = amn
∴ (24)2 = 28
Exercise 11.3 Solutions
Q 1. Express the number appearing in the following statements in standard form.
- The distance between Earth and the Moon is 384,000,000 m.
- The speed of light in a vacuum is 300,000,000 m/s.
Solutions: The solutions are given below.
- The standard form of the given number is 3.84 × 108 m.
- The standard form of the given number is 3 × 108 m/s.
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FAQs
Ans: The concept of exponents can be explained by using an example. The number 104 is read as 10 raised to the power of 4 or simply as the fourth power of 10. Here, 104 is called the exponential form of 10,000.
Ans: A Monomial is an algebraic expression with only one term. For example, 7𝑥y, – 5m, 3z2, 4 etc.
Ans: A Polynomial is an expression with one or more terms. Thus, a monomial, a binomial and a trinomial are all polynomials.
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This was all about NCERT Solutions Class 7 Maths Chapter 11, Exponents and Powers in which we studied the terms, factors, and coefficients of Exponents and Powers. Download the NCERT Solutions Class 7 Maths Chapter 11 Notes and Solutions PDF to ace your e𝑥am preparations. Follow the CBSE Class 7 Maths Solutions and Notes for more such chapter notes and important questions and answers for preparation for CBSE Class 7 Maths.