Exercise 4.1 — Laws of Exponents

What This Exercise Covers

Exercise 4.1 from Chapter 4 — Exponents and Powers gives Class 8 students systematic practice applying all the core laws of exponents. The questions move from straightforward simplification to multi-step problems involving negative exponents, fractional bases, and finding unknown values. This exercise appears in the CBSE, Telangana, and Andhra Pradesh Class 8 Mathematics syllabi and forms the foundation for working with standard form and algebraic expressions in later chapters.

Laws of Exponents — Your Toolkit for This Exercise

Every question in Exercise 4.1 can be solved by applying one or more of these rules. Knowing which law to reach for is the key skill being built here.

• Product Rule — aᵐ × aⁿ = aᵐ⁺ⁿ: add exponents when multiplying the same base
• Quotient Rule — aᵐ ÷ aⁿ = aᵐ⁻ⁿ: subtract exponents when dividing the same base
• Power of a Power — (aᵐ)ⁿ = aᵐⁿ: multiply the exponents
• Product of Powers — aᵐ × bᵐ = (ab)ᵐ: combine bases when the exponent is the same
• Zero Exponent — a⁰ = 1 for any non-zero a
• Negative Exponent — a⁻ⁿ = 1/aⁿ: a negative exponent means take the reciprocal
• Equal Bases Rule — if aᵐ = aⁿ, then m = n: equate exponents when bases are the same

Worked Examples — Step by Step

Here are representative problems from each question type, solved with the reasoning explained at each step.

Simplifying a negative exponent (Q1): To find the value of 4⁻³, apply a⁻ⁿ = 1/aⁿ to rewrite it as 1/4³ = 1/64. For a fractional base like (3/4)⁻³, flipping the fraction gives (4/3)³ = 64/27.

Combining the product rule (Q2): For 12⁴ × 12⁵ × 12⁶, since all three terms share the same base, simply add the exponents: 4 + 5 + 6 = 15, giving 12¹⁵. For a mixed case like (−3)⁴ × 7⁴, the exponents are equal across different bases, so use aᵐ × bᵐ = (ab)ᵐ to get (−21)⁴.

Multi-law simplification (Q4): An expression like 4⁰ + 5⁻¹ × 5² × (1/3) requires handling several rules at once — replacing 4⁰ with 1, converting 5⁻¹ to 1/5, then applying the quotient rule on 5² ÷ 5¹ to get 5¹ = 5, and finally multiplying to reach 10. Working methodically, one rule at a time, is the key habit to develop.

Finding the Unknown — Questions 6 and 7

Questions 6 and 7 introduce a different challenge: rather than evaluating an expression, you find a missing value of n or x. The strategy is always to simplify both sides until they have the same base, then equate the exponents using the Equal Bases Rule.

For example, in (2/3)³ × (2/3)⁵ = (2/3)ⁿ⁻², apply the product rule on the left to get (2/3)⁸, then set 8 = n − 2 to find n = 10. Similarly, to find x in 2⁻³ = (1/2)ˣ, rewrite 1/2ˣ as 2⁻ˣ, then match exponents to get x = 3.

Substitution Problems — Question 9

Question 9 asks you to evaluate expressions like 9m² − 10n³ when m = 3 and n = 2. The important point here is to respect the order of operations: compute the powers first (3² = 9, 2³ = 8), then multiply by coefficients, and finally add or subtract. For instance, 9(9) − 10(8) = 81 − 80 = 1. Rushing this step and computing 9 × 3 instead of 9 × 3² is the most common error.

Common Mistakes to Avoid

• Applying the product rule across different bases — aᵐ × bⁿ cannot be simplified this way; the bases must match.
• Confusing (−2)⁷ with −2⁷. The brackets matter: (−2)⁷ = −128, while −2⁷ means −(2⁷) = −128 here by coincidence, but (−2)⁴ = +16 while −2⁴ = −16.
• Forgetting that a⁰ = 1, not 0 — this trips up many students in multi-term expressions.
• When flipping a negative exponent on a fraction, forgetting to flip the entire fraction: (3/4)⁻² becomes (4/3)², not 4/3².

What This Exercise Prepares You For

Mastering Exercise 4.1 sets you up for Exercise 4.2 on Standard Form, where these same laws are used to express and compare very large and very small numbers. The substitution skills from Question 9 also directly support Algebraic Expressions. In Class 9 and 10, exponent laws reappear in Real Numbers and throughout polynomial operations.