10.1 Introduction | Class 8 Mathematics Notes
By ConceptScroll Team · Published on 17 July 2026 · 3 min read
10.1 Introduction – this guide gives you a concise, exam-ready overview of 10.1 Introduction from Class 8 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
10.3 Laws of Exponents
This section discusses the fundamental laws governing exponents, extending them to include negative and zero exponents. The laws provide rules for multiplying, dividing, and raising powers to powers, which simplify algebraic expressions involving exponents. The key laws are: (i) a^m × a^n = a^(m + n), (ii) a^m ÷ a^n = a^(m - n), (iii) (a^m)^n = a^(m × n), (iv) a^m × b^m = (ab)^m, (v) (a^m) ÷ (b^m) = (a/b)^m, and (vi) a^0 = 1 for any non-zero a. These laws hold for any non-zero integer base a and integers m and n (positive, negative, or zero). The section provides detailed examples demonstrating these laws, such as simplifying expressions like (-4)^5 × (-4)^-10 = (-4)^-5 = 1/(-4)^5, and 2^5 ÷ 2^-6 = 2^(5 + 6) = 2^11. It also shows how to express powers with different bases, for example, 4^-3 = (2^2)^-3 = 2^-6. Understanding these laws is crucial for simplifying exponential expressions and solving algebraic problems efficiently.
📊 Diagram: Figures showing examples of laws of exponents with negative and positive powers; step-by-step simplifications of expressions.
🧪 Activity: Simplify given expressions using laws of exponents, including negative powers.
🔗 Connection: Leads to worked examples applying these laws and further practice problems.
Frequently asked questions
1. Evaluate. (i) $ 3^{-2} $ (ii) $ (-4)^{-2} $ (iii) $ \left( \frac{1}{2} \right)^{-5} $
Solution: (i) $3^{-2} = \frac{1}{3^2} = \frac{1}{9}$
(ii) $(-4)^{-2} = \frac{1}{(-4)^2} = \frac{1}{16}$
(iii) $\left( \frac{1}{2} \right)^{-5} = 2^5 = 32$
2. Simplify and express the result in power notation with positive exponent. (i) $ (-4)^5 \div (-4)^8 $ (ii) $ \left( \frac{1}{2^3} \right)^2 $ (iii) $ (-3)^4 \times \left( \frac{5}{3} \right)^4 $ (iv) $ (3^{-7} \div 3^{-10}) \times 3^{-5} $ (v) $ 2^{-3} \times (-7)^{-3} $
Solution: (i) $(-4)^5 \div (-4)^8 = (-4)^{5-8} = (-4)^{-3} = \frac{1}{(-4)^3} = \frac{1}{-64} = -\frac{1}{64}$
(ii) $\left( \frac{1}{2^3} \right)^2 = \left( 2^{-3} \right)^2 = 2^{-6} = \frac{1}{2^6} = \frac{1}{64}$
(iii) $(-3)^4 \times \left( \frac{5}{3} \right)^4 = \left( -3 \times \frac{5}{3} \right)^4 = (-5)^4 = 5^4 = 625$
(iv) $(3^{-7} \div 3^{-10}) \times 3^{-5} = 3^{-7 - (-10)} \times 3^{-5} = 3^{3} \times 3^{-5} = 3^{3-5} = 3^{-2} = \frac{1}{9}$
(v) $2^{-3} \times (-7)^{-3} = \frac{1}
3. Find the value of. (i) $ (3^0 + 4^{-1}) \times 2^2 $ (ii) $ (2^{-1} \times 4^{-1}) \div 2^{-2} $ (iii) $ \left( \frac{1}{2} \right)^{-2} + \left( \frac{1}{3} \right)^{-2} + \left( \frac{1}{4} \right)^{-2} $ (iv) $ (3^{-1} + 4^{-1} + 5^{-1})^0 $ (v) $ \left\{ \left( \frac{-2}{3} \right)^{-2} \right\}^2 $
Solution: (i) $3^0 = 1$, $4^{-1} = \frac{1}{4}$, so $(1 + \frac{1}{4}) \times 2^2 = \frac{5}{4} \times 4 = 5$
(ii) $(2^{-1} \times 4^{-1}) \div 2^{-2} = \left( \frac{1}{2} \times \frac{1}{4} \right) \div \frac{1}{2^2} = \frac{1}{8} \div \frac{1}{4} = \frac{1}{8} \times 4 = \frac{1}{2}$
(iii) $\left( \frac{1}{2} \right)^{-2} = 2^2 = 4$, $\left( \frac{1}{3} \right)^{-2} = 3^2 = 9$, $\left( \frac{1}{4} \right)^{-2} = 4^2 = 16$, sum = $4 + 9 + 16 = 29$
(iv) Any number to the power 0 is 1, so $(3^
4. Evaluate (i) $\frac{8^{-1} \times 5^3}{2^{-4}}$ (ii) $(5^{-1} \times 2^{-1}) \times 6^{-1}$
Solution: (i) $\frac{8^{-1} \times 5^3}{2^{-4}} = \frac{\frac{1}{8} \times 125}{\frac{1}{2^{-4}}} = \frac{\frac{125}{8}}{2^{-4}} = \frac{125}{8} \times 2^{4} = \frac{125}{8} \times 16 = 125 \times 2 = 250$
(ii) $(5^{-1} \times 2^{-1}) \times 6^{-1} = \frac{1}{5} \times \frac{1}{2} \times \frac{1}{6} = \frac{1}{60}$
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