CBSE Class 8 Maths Chapter 12: Exponents and Powers NCERT Solutions

NCERT Solutions PDF Class 8 PDF

CBSE Class 8 Mathematics Chapter 12, Exponents and Powers, delves into representing numbers efficiently using exponents. This chapter is key to understanding how to work with very large or very small numbers. The NCERT Solutions offer detailed, step-by-step guidance through exercises focused on negative and fractional exponents, as well as the fundamental laws governing them. Students will practice simplifying expressions, evaluating powers, and converting between standard and exponential forms. Mastering these concepts is essential for building a solid algebraic foundation, which is crucial for success in higher mathematics and competitive exams. The provided solutions aim to demystify complex problems, ensuring a clear understanding of each principle and enabling students to confidently tackle any question related to exponents and powers.

Quick info

BoardCBSE
ClassClass 8
SubjectMathematics
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 12: Exponents and Powers

Chapter summary

Chapter 12, Exponents and Powers, for Class 8 Mathematics focuses on understanding and applying the rules of exponents. The NCERT Solutions cover evaluating expressions with positive and negative exponents, simplifying terms using exponent laws (product, quotient, power of a power), and working with fractional bases. The exercises involve calculations and expressing results in specific forms, reinforcing the practical application of exponent rules.

Learning outcomes

  • Understand the concept of exponents and powers.
  • Apply the laws of exponents to simplify expressions.
  • Evaluate expressions involving positive and negative exponents.
  • Solve problems involving fractional bases and powers.
  • Express results in power notation with positive exponents.

Topics covered

Paper topics

  • Exponents and Powers
  • Positive Exponents
  • Negative Exponents
  • Laws of Exponents
  • Product of Powers
  • Quotient of Powers
  • Power of a Power
  • Powers with Fractional Bases
  • Evaluating Expressions
  • Simplifying Expressions
  • Standard Form
  • Zero Exponent

Important topics

  • Laws of Exponents (Product, Quotient, Power of a Power)
  • Handling Negative Exponents
  • Evaluating Expressions with Mixed Operations
  • Simplifying Complex Exponent Terms
  • Understanding $a^0 = 1$

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Questions and Solutions

Exercise 12.1, Question 1

Evaluate:
  1. 3^{-2}
  2. (-4)^{-2}
  3. \left(\frac{1}{2}\right)^{-5}
Solution:

We will evaluate each part using the law of exponents $a^{-m} = \frac{1}{a^m}$.

  1. For 3^{-2}, we apply the rule:

    3^{-2} = \frac{1}{3^2}

    Now, we calculate the square of 3:

    = \frac{1}{9}

    Thus, the value of 3^{-2} is \frac{1}{9}.

  2. For (-4)^{-2}, we apply the same rule:

    \left(-4\right)^{-2} = \frac{1}{\left(-4\right)^2}

    Next, we calculate the square of -4:

    = \frac{1}{16}

    Therefore, the value of (-4)^{-2} is \frac{1}{16}.

  3. For \left(\frac{1}{2}\right)^{-5}, we can use the rule $(\frac{a}{b})^{-m} = (\frac{b}{a})^m$.

    \left(\frac{1}{2}\right)^{-5} = \left(\frac{2}{1}\right)^5

    This simplifies to:

    = (2)^5

    Calculating the fifth power of 2:

    = 32

    Hence, the value of \left(\frac{1}{2}\right)^{-5} is 32.

Exercise 12.1, Question 2

Simplify and express the result in power notation with positive exponent:
  1. (-4)^5 \div (-4)^8
  2. \left(\frac{1}{2^3}\right)^2
  3. \left(-3\right)^4 \times \left(\frac{5}{3}\right)^4
  4. (3^{-7} \div 3^{-10}) \times 3^{-5}
  5. 2^{-3} \times (-7)^{-3}
Solution:

We will use the laws of exponents to simplify each expression.

  1. To simplify (-4)^5 \div (-4)^8, we use the quotient rule $a^m \div a^n = a^{m-n}$:

    (-4)^5 \div (-4)^8 = (-4)^{5-8}

    = (-4)^{-3}

    To express this with a positive exponent, we use the rule $a^{-m} = \frac{1}{a^m}$:

    = \frac{1}{(-4)^3}

    The result in power notation with a positive exponent is (-4)^{-3} or \frac{1}{(-4)^3}.

  2. To simplify \left(\frac{1}{2^3}\right)^2, we use the power of a power rule $(a^m)^n = a^{m \times n}$ and the rule $(\frac{a}{b})^m = \frac{a^m}{b^m}$:

    \left(\frac{1}{2^3}\right)^2 = \frac{1^2}{\left(2^3\right)^2}

    Applying the power of a power rule to the denominator:

    = \frac{1}{2^{3 \times 2}} = \frac{1}{2^6}

    The simplified expression with a positive exponent is \frac{1}{2^6}.

  3. To simplify \left(-3\right)^4 \times \left(\frac{5}{3}\right)^4, we use the rule $(ab)^m = a^m b^m$ and $(\frac{a}{b})^m = \frac{a^m}{b^m}$:

    \left(-3\right)^4 \times \left(\frac{5}{3}\right)^4 = (-3)^4 \times \frac{5^4}{3^4}

    We can write $(-3)^4$ as $(1 \times 3)^4 = 1^4 \times 3^4 = 3^4$ since the exponent is even.

    = 3^4 \times \frac{5^4}{3^4}

    Now, we cancel out the 3^4 terms:

    = 5^4

    The simplified result is 5^4.

  4. To simplify (3^{-7} \div 3^{-10}) \times 3^{-5}, we first use the quotient rule $a^m \div a^n = a^{m-n}$:

    (3^{-7} \div 3^{-10}) = 3^{-7 - (-10)} = 3^{-7 + 10} = 3^3

    Now, we multiply this result by 3^{-5} using the product rule $a^m \times a^n = a^{m+n}$:

    3^3 \times 3^{-5} = 3^{3 + (-5)} = 3^{3-5} = 3^{-2}

    To express this with a positive exponent, we use $a^{-m} = \frac{1}{a^m}$:

    = \frac{1}{3^2}

    The simplified result with a positive exponent is \frac{1}{3^2}.

  5. To simplify 2^{-3} \times (-7)^{-3}, we use the rule $(ab)^m = a^m b^m$:

    2^{-3} \times (-7)^{-3} = \{2 \times (-7)\}^{-3}

    = (-14)^{-3}

    To express this with a positive exponent, we use $a^{-m} = \frac{1}{a^m}$:

    = \frac{1}{(-14)^3}

    The simplified result with a positive exponent is \frac{1}{(-14)^3}.

Exercise 12.1, Question 3

Find the value of:
  1. (3^0 + 4^{-1}) \times 2^2
  2. (2^{-1} \times 4^{-1}) \div 2^{-2}
  3. \left(\frac{1}{2}\right)^{-2} + \left(\frac{1}{3}\right)^{-2} + \left(\frac{1}{4}\right)^{-2}
  4. \left(3^{-1} + 4^{-1} + 5^{-1}\right)^0
  5. \left\{\left(\frac{-2}{3}\right)^{-2}\right\}^2
Solution:

We will evaluate each expression using the laws of exponents.

  1. For (3^0 + 4^{-1}) \times 2^2:

    First, we know that any non-zero number raised to the power of 0 is 1, so 3^0 = 1. Also, 4^{-1} = \frac{1}{4} and 2^2 = 4.

    \left(1 + \frac{1}{4}\right) \times 4

    Combine the terms inside the parenthesis:

    \left(\frac{4}{4} + \frac{1}{4}\right) \times 4 = \frac{5}{4} \times 4

    Now, multiply: = 5 The value is 5.

  2. For (2^{-1} \times 4^{-1}) \div 2^{-2}:

    We can rewrite 4^{-1} as (2^2)^{-1} = 2^{-2}.

    \left(2^{-1} \times 2^{-2}\right) \div 2^{-2}

    Using the product rule $a^m \times a^n = a^{m+n}$ for the terms in the parenthesis:

    = 2^{-1 + (-2)} \div 2^{-2} = 2^{-3} \div 2^{-2}

    Using the quotient rule $a^m \div a^n = a^{m-n}$:

    = 2^{-3 - (-2)} = 2^{-3 + 2} = 2^{-1}

    Expressing with a positive exponent:

    = \frac{1}{2}

    The value is \frac{1}{2}.

  3. For \left(\frac{1}{2}\right)^{-2} + \left(\frac{1}{3}\right)^{-2} + \left(\frac{1}{4}\right)^{-2}:

    We use the rule $(\frac{a}{b})^{-m} = (\frac{b}{a})^m$ for each term:

    \left(\frac{2}{1}\right)^2 + \left(\frac{3}{1}\right)^2 + \left(\frac{4}{1}\right)^2

    = 2^2 + 3^2 + 4^2

    Calculate the squares:

    = 4 + 9 + 16

    Add the results:

    = 29

    The value is 29.

  4. For \left(3^{-1} + 4^{-1} + 5^{-1}\right)^0:

    Any non-zero number raised to the power of 0 is 1. Since the base (3^{-1} + 4^{-1} + 5^{-1}) is a sum of positive numbers, it will be non-zero.

    \left(3^{-1} + 4^{-1} + 5^{-1}\right)^0 = 1

    The value is 1.

  5. For \left\{\left(\frac{-2}{3}\right)^{-2}\right\}^2:

    First, we simplify the inner part \left(\frac{-2}{3}\right)^{-2} using the rule $(\frac{a}{b})^{-m} = (\frac{b}{a})^m$:

    \left(\frac{-2}{3}\right)^{-2} = \left(\frac{3}{-2}\right)^2 = \frac{3^2}{(-2)^2} = \frac{9}{4}

    Now, we raise this result to the power of 2:

    \left(\frac{9}{4}\right)^2 = \frac{9^2}{4^2} = \frac{81}{16}

    The value is \frac{81}{16}.

Common mistakes

  • Incorrectly applying the rule for negative exponents (e.g., confusing $a^{-m}$ with $-a^m$).
  • Errors in simplifying expressions with multiple exponent rules applied together.
  • Mistakes in handling signs when dealing with negative bases and exponents.
  • Incorrectly applying the power of a power rule, such as adding instead of multiplying exponents.

Revision tips

  • Memorize all the laws of exponents and practice applying them to different types of problems.
  • Pay close attention to the signs of the base and the exponent, especially with negative numbers.
  • Work through each example and exercise problem step-by-step, ensuring each step is logical.
  • Try to solve problems without looking at the solution first, then check your work.

Practice MCQs

Q1. What is the value of $3^{-2}$?

Q2. Simplify: $(-4)^5 \div (-4)^8$

Q3. What is the value of $(2^{-1} \times 4^{-1}) \div 2^{-2}$?

Q4. What is the value of $(3^0 + 4^{-1}) \times 2^2$?

Q5. Simplify: $(\frac{1}{2})^{-2} + (\frac{1}{3})^{-2} + (\frac{1}{4})^{-2}$

Q6. What is the value of $(3^{-1} + 4^{-1} + 5^{-1})^0$?

Q7. Simplify: $\{(\frac{-2}{3})^{-2}\}^2$

Frequently asked questions

What is the main concept covered in CBSE Class 8 Maths Chapter 12?

Chapter 12, Exponents and Powers, focuses on understanding and applying the rules of exponents to simplify and evaluate mathematical expressions, including those with positive, negative, and zero exponents.

How do these NCERT Solutions help students?

These solutions provide clear, step-by-step explanations for each problem in Exercise 12.1, helping students understand the application of exponent laws and build confidence in solving similar problems.

What is the rule for negative exponents?

The rule for negative exponents states that $a^{-m} = \frac{1}{a^m}$, where 'a' is any non-zero number and 'm' is a positive integer. This means a negative exponent in the numerator becomes a positive exponent in the denominator, and vice versa.

How can I simplify expressions with exponents?

You can simplify expressions by applying the laws of exponents, such as $a^m \times a^n = a^{m+n}$, $a^m \div a^n = a^{m-n}$, and $(a^m)^n = a^{m \times n}$. Remember to also handle negative exponents and fractional bases correctly.

What is the value of any non-zero number raised to the power of zero?

Any non-zero number raised to the power of zero is always equal to 1. For example, $5^0 = 1$, $(-7)^0 = 1$, and $(\frac{2}{3})^0 = 1$.

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