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

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CBSE Class 8 Mathematics, Chapter 12, Exponents and Powers, delves into the fundamental concepts of working with exponents. This chapter equips students with the ability to evaluate expressions involving negative exponents and simplify complex expressions by applying the laws of exponents. Key rules such as the product rule ($a^m \times a^n = a^{m+n}$), the quotient rule ($a^m \div a^n = a^{m-n}$), and the rule for negative exponents ($a^{-m} = \frac{1}{a^m}$) are thoroughly explained and demonstrated. The solutions provided offer a step-by-step approach to understanding and solving various problems related to powers. This resource is designed to build a strong foundation in this area, ensuring students can confidently tackle related questions in their examinations and future mathematical studies.

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Quick info

BoardCBSE
ClassClass 8
SubjectMathematics
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 12

Chapter summary

Chapter 12, Exponents and Powers, for Class 8 Maths NCERT Solutions focuses on understanding and applying the rules of exponents. This chapter's solutions cover evaluating expressions with positive and negative exponents, simplifying complex expressions using laws of exponents, and expressing results in a specific power notation. Key concepts include the reciprocal rule for negative exponents and combining terms with the same base or exponent.

Learning outcomes

  • Understand and apply the rule $a^{-m} = \frac{1}{a^m}$ to evaluate expressions.
  • Simplify expressions involving division and multiplication of powers with the same base.
  • Apply the rule $(\frac{a}{b})^m = \frac{a^m}{b^m}$ and $(ab)^m = a^m b^m$ for simplification.
  • Evaluate expressions with zero exponents using the rule $a^0 = 1$.
  • Solve problems involving nested exponents using the rule $(a^m)^n = a^{mn}$.
  • Express results in power notation with positive exponents.

Topics covered

Paper topics

  • Laws of Exponents
  • Positive Exponents
  • Negative Exponents
  • Zero Exponent
  • Evaluating Expressions
  • Simplifying Expressions
  • Power Notation
  • Reciprocals

Important topics

  • Laws of Exponents (multiplication, division, power of a power)
  • Handling Negative Exponents
  • Evaluating Expressions with Zero Exponents

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Questions and Solutions

Question 1

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

We use the law of exponents $a^{-m} = \frac{1}{a^m}$ to evaluate these expressions.

  1. For 3^{-2}:

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

    Calculating the square of 3:

    \frac{1}{3^2} = \frac{1}{9}

    Thus, 3^{-2} = \frac{1}{9}.

  2. For (-4)^{-2}:

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

    Calculating the square of -4:

    \frac{1}{(-4)^2} = \frac{1}{16}

    Thus, (-4)^{-2} = \frac{1}{16}.

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

    We can use the rule $\left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^{m}$.

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

    Now, we calculate the fifth power of 2:

    \left(\frac{2}{1}\right)^{5} = 2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32

    Thus, \left(\frac{1}{2}\right)^{-5} = 32.

Question 2

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

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

    (-4)^5 \div (-4)^8 = (-4)^{5-8} = (-4)^{-3}

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

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

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

  2. Simplify (\frac{1}{2^3})^2:

    Using the law $(\frac{a}{b})^m = \frac{a^m}{b^m}$ and $(a^m)^n = a^{mn}$:

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

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

    The result is \frac{1}{2^6}.

  3. Simplify \left(-3\right)^4 \times \left(\frac{5}{3}\right)^4:

    Using the law $(ab)^m = a^m b^m$ 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 know that $(-3)^4 = ((-1) \times 3)^4 = (-1)^4 \times 3^4 = 1 \times 3^4 = 3^4$.

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

    Now, cancel out 3^4:

    \frac{3^4}{3^4} \times 5^4 = 3^{4-4} \times 5^4 = 3^0 \times 5^4

    Since $a^0 = 1$:

    1 \times 5^4 = 5^4

    The result is 5^4.

  4. Simplify (3^{-7} \div 3^{-10}) \times 3^{-5}:

    First, simplify the division using $a^m \div a^n = a^{m-n}$:

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

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

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

    To express this with a positive exponent:

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

    The result is \frac{1}{3^2}.

  5. Simplify 2^{-3} \times (-7)^{-3}:

    Using the law $(ab)^m = a^m b^m$:

    2^{-3} \times (-7)^{-3} = \{2 \times (-7)\}^{-3} = (-14)^{-3}

    To express this with a positive exponent:

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

    The result is \frac{1}{(-14)^3}.

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

    We know that $a^0 = 1$ and $a^{-m} = \frac{1}{a^m}$.

    (3^0 + 4^{-1}) \times 2^2 = (1 + \frac{1}{4^1}) \times 2^2

    = (1 + \frac{1}{4}) \times 4

    Combine the terms inside the parenthesis:

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

    Now, multiply:

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

    The value is 5.

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

    Using $a^{-m} = \frac{1}{a^m}$ and $(ab)^m = a^m b^m$:

    (2^{-1} \times 4^{-1}) = (2 \times 4)^{-1} = 8^{-1} = \frac{1}{8}

    And $2^{-2} = \frac{1}{2^2} = \frac{1}{4}$.

    So the expression becomes:

    \frac{1}{8} \div \frac{1}{4}

    Dividing by a fraction is the same as multiplying by its reciprocal:

    \frac{1}{8} \times \frac{4}{1} = \frac{4}{8} = \frac{1}{2}

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

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

    Using the rule $\left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^{m}$:

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

    \left(\frac{1}{3}\right)^{-2} = \left(\frac{3}{1}\right)^{2} = 3^2 = 9

    \left(\frac{1}{4}\right)^{-2} = \left(\frac{4}{1}\right)^{2} = 4^2 = 16

    Now, add these values:

    4 + 9 + 16 = 29

    The value is 29.

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

    Any non-zero number raised to the power of 0 is 1.

    The expression inside the parenthesis is 3^{-1} + 4^{-1} + 5^{-1} = \frac{1}{3} + \frac{1}{4} + \frac{1}{5}. This sum is clearly not zero.

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

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

    First, simplify the inner exponent using $\left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^{m}$:

    \left(\frac{-2}{3}\right)^{-2} = \left(\frac{3}{-2}\right)^{2} = \left(\frac{3}{2}\right)^{2} (since $(-2)^2 = 2^2$)

    Now, apply the outer exponent:

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

    Using the law $(a^m)^n = a^{mn}$:

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

    Now, evaluate this:

    \left(\frac{3}{2}\right)^{4} = \frac{3^4}{2^4} = \frac{81}{16}

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

Common mistakes

  • Incorrectly applying the rule for negative exponents, e.g., writing $a^{-m} = -a^m$ instead of $\frac{1}{a^m}$.
  • Errors in exponent arithmetic, especially when dealing with subtraction or addition of exponents during division or multiplication.
  • Forgetting to apply the exponent to both the numerator and denominator when dealing with fractions raised to a power.
  • Misinterpreting the order of operations when simplifying complex expressions with multiple exponents.

Revision tips

  • Practice applying the laws of exponents ($a^m \times a^n$, $a^m \div a^n$, $(a^m)^n$, $(ab)^m$, $(\frac{a}{b})^m$) consistently.
  • Pay close attention to the signs of exponents and bases when performing calculations.
  • Work through each step of the simplification process methodically to avoid arithmetic errors.
  • Review the special cases for zero and negative exponents thoroughly.

Practice MCQs

Q1. What is the value of $5^{-3}$?

Q2. Simplify: $2^5 \div 2^2$

Q3. What is the value of $(7^0)$?

Q4. Simplify: $(\frac{3}{4})^2$

Q5. Evaluate: $3^{-2} \times 3^4$

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 laws of exponents to simplify and evaluate numerical expressions involving positive, negative, and zero exponents.

How are negative exponents handled in this chapter?

Negative exponents are handled using the rule $a^{-m} = \frac{1}{a^m}$, which means the expression is the reciprocal of the base raised to the positive exponent.

What are the key laws of exponents discussed?

The key laws include $a^m \times a^n = a^{m+n}$, $a^m \div a^n = a^{m-n}$, $(a^m)^n = a^{mn}$, $(ab)^m = a^m b^m$, and $(\frac{a}{b})^m = \frac{a^m}{b^m}$, along with the rule for zero exponent, $a^0 = 1$.

How do these NCERT solutions help students?

These solutions provide clear, step-by-step explanations for solving problems related to exponents and powers, helping students grasp the concepts and practice applying the exponent rules accurately for better exam preparation.

What is the value of any non-zero number raised to the power of 0?

Any non-zero number raised to the power of 0 is always equal to 1. For example, $10^0 = 1$, $(-5)^0 = 1$, and $(\frac{2}{3})^0 = 1$.

Important links

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