What is 2¹⁰⁰ ÷ 2²⁵, written as a power of 2?
Hint. Use the rule nᵃ ÷ nᵇ = nᵃ⁻ᵇ for same-base division.
Step 1 — Apply the same-base division rule. 2¹⁰⁰ ÷ 2²⁵ = 2¹⁰⁰⁻²⁵.
Step 2 — Subtract the exponents. 100 − 25 = 75.
✦ Answer: 2⁷⁵.
Where students slip. Dividing the exponents (100 ÷ 25 = 4) instead of subtracting them — same-base division subtracts exponents, it never divides them.
In the rule xᵃ ÷ xᵃ = x⁰ = 1 (x ≠ 0), why is the condition x ≠ 0 needed — why can't the base be 0?
Hint. Try substituting x = 0 into the same rule and see what expression results.
Step 1 — Substitute x = 0 into the rule. 0ᵃ ÷ 0ᵃ would become 0⁰.
Step 2 — Recall what 0⁰ means. 0⁰ has no defined value (dividing 0 by 0 is undefined), so the rule x⁰ = 1 cannot be applied when x = 0.
✦ Answer: If the base were 0, the rule would require 0⁰, which is not defined — so x must be a nonzero number for x⁰ = 1 to make sense.
Where students slip. Assuming x⁰ = 1 holds for every number including 0 — the rule specifically excludes a base of 0, since 0⁰ is undefined.
The laws of exponents were shown using a and b as counting numbers. Can a and b be any integers, and will the generalised forms still hold true?
Hint. Check whether the earlier examples with negative exponents (like 2⁻¹, 2⁻⁶) fit the same pattern as the positive-exponent cases.
Step 1 — Recall the earlier negative-exponent results. 2⁴ ÷ 2⁵ = 2⁻¹ and 2⁴ ÷ 2¹⁰ = 2⁻⁶ both used the same subtraction-of-exponents rule as the positive-exponent cases.
Step 2 — Confirm the pattern extends. Since these negative-exponent cases followed the identical rule nᵃ ÷ nᵇ = nᵃ⁻ᵇ without any change in form, the generalised laws hold for all integers, not just counting numbers.
✦ Answer: Yes — a and b can be any integers, and the same laws of exponents continue to hold true.
Where students slip. Assuming the laws of exponents need separate versions for negative exponents — the same rules apply unchanged, once negative exponents are defined as reciprocals of positive ones.
Write equivalent forms of the following: (i) 2⁻⁴ (ii) 10⁻⁵ (iii) (−7)⁻² (iv) (−5)⁻³ (v) 10⁻¹⁰⁰
Hint. A negative exponent means 'reciprocal of the positive-exponent form' — n⁻ᵃ = 1/nᵃ.
Step 1 — Apply n⁻ᵃ = 1/nᵃ to each. (i) 2⁻⁴ = 1/2⁴. (ii) 10⁻⁵ = 1/10⁵. (iii) (−7)⁻² = 1/(−7)². (iv) (−5)⁻³ = 1/(−5)³. (v) 10⁻¹⁰⁰ = 1/10¹⁰⁰, since a negative exponent always means the reciprocal of the same power with a positive exponent.
✦ Answer: (i) 1/2⁴ (ii) 1/10⁵ (iii) 1/(−7)² (iv) 1/(−5)³ (v) 1/10¹⁰⁰.
Where students slip. Dropping the negative sign inside the base when rewriting (iii) and (iv) — the base (including its sign) stays exactly as given; only the exponent's sign flips to move it into the denominator.
Simplify and write the answers in exponential form: (i) 2⁻⁴ × 2⁷ (ii) 3² × 3⁻⁵ × 3⁶ (iii) p³ × p⁻¹⁰ (iv) 2⁴ × (−4)⁻² (v) 8ᵖ × 8ᑫ
Hint. For same-base multiplication, add the exponents — this works the same way whether the exponents are positive or negative.
Step 1 — (i) Add exponents: −4 + 7 = 3, so 2⁻⁴ × 2⁷ = 2³.
Step 2 — (ii) Add exponents: 2 + (−5) + 6 = 3, so 3² × 3⁻⁵ × 3⁶ = 3³.
Step 3 — (iii) Add exponents: 3 + (−10) = −7, so p³ × p⁻¹⁰ = p⁻⁷.
Step 4 — (iv) Rewrite (−4)⁻² as 1/(−4)² = 1/16, and 2⁴ = 16, so the product is 16 × (1/16) = 1.
Step 5 — (v) Add exponents: 8ᵖ × 8ᑫ = 8ᵖ⁺ᑫ.
✦ Answer: (i) 2³ (ii) 3³ (iii) p⁻⁷ (iv) 1 (v) 8ᵖ⁺ᑫ.
Where students slip. Treating (iv) as if it doesn't simplify neatly — converting (−4)⁻² to a fraction first shows it exactly cancels 2⁴, leaving a clean answer of 1.
How many times larger than 4⁻² is 4²?
Hint. Divide 4² by 4⁻² using the same-base division rule.
Step 1 — Set up the division. 4² ÷ 4⁻² = 4²⁻⁽⁻²⁾ = 4⁴, since subtracting a negative exponent means adding it.
Step 2 — Evaluate. 4⁴ = 256.
✦ Answer: 4² is 4⁴ (= 256) times larger than 4⁻².
Where students slip. Subtracting the exponents as 2 − 2 = 0 — since the second exponent is negative, subtracting it means adding 2, giving 4 (not 0) as the resulting exponent.
Using the powers of 7 (7⁰=1, 7¹=7, 7²=49, 7³=343, 7⁴=2401, 7⁵=16807, 7⁶=117649, 7⁷=823543, and 7⁻¹=1/7, 7⁻²=1/49, 7⁻³=1/343), work out: (i) 2401 × 49 (ii) 49³ (iii) 343 × 2401 (iv) 16807 ÷ 49 (v) 7 ÷ 343 (vi) 16807 ÷ 823543 (vii) 117649 × (1/343) (viii) (1/343) × (1/343) — expressing each as a power of 7.
Hint. Convert every number to its power-of-7 form first, then just add or subtract exponents.
Step 1 — (i) 2401 × 49 = 7⁴ × 7² = 7⁶.
Step 2 — (ii) 49³ = (7²)³ = 7⁶.
Step 3 — (iii) 343 × 2401 = 7³ × 7⁴ = 7⁷.
Step 4 — (iv) 16807 ÷ 49 = 7⁵ ÷ 7² = 7³.
Step 5 — (v) 7 ÷ 343 = 7¹ ÷ 7³ = 7⁻².
Step 6 — (vi) 16807 ÷ 823543 = 7⁵ ÷ 7⁷ = 7⁻².
Step 7 — (vii) 117649 × (1/343) = 7⁶ × 7⁻³ = 7³.
Step 8 — (viii) (1/343) × (1/343) = 7⁻³ × 7⁻³ = 7⁻⁶.
✦ Answer: (i) 7⁶ (ii) 7⁶ (iii) 7⁷ (iv) 7³ (v) 7⁻² (vi) 7⁻² (vii) 7³ (viii) 7⁻⁶ — since converting every term to a power of 7 turns each calculation into simple exponent addition or subtraction.
Where students slip. Multiplying out the actual large numbers instead of converting to powers of 7 first — once every term is written as 7 to some exponent, the arithmetic collapses to adding or subtracting small integers.
Write these numbers using powers of 10, the way 47561 = (4 × 10⁴) + (7 × 10³) + (5 × 10²) + (6 × 10¹) + (1 × 10⁰): (i) 172 (ii) 5642 (iii) 6374
Hint. Match each digit to its place value, from the highest power of 10 down to 10⁰.
Step 1 — Expand 172. 172 = (1 × 10²) + (7 × 10¹) + (2 × 10⁰).
Step 2 — Expand 5642. 5642 = (5 × 10³) + (6 × 10²) + (4 × 10¹) + (2 × 10⁰).
Step 3 — Expand 6374. 6374 = (6 × 10³) + (3 × 10²) + (7 × 10¹) + (4 × 10⁰), since each digit is exactly one power of 10 higher than the digit to its right.
✦ Answer: 172 = 1×10² + 7×10¹ + 2×10⁰; 5642 = 5×10³ + 6×10² + 4×10¹ + 2×10⁰; 6374 = 6×10³ + 3×10² + 7×10¹ + 4×10⁰.
Where students slip. Misaligning a digit with the wrong power of 10 — the units digit is always ×10⁰, and the power increases by one for each place moved left.
Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp102.pdf). Questions are scattered as 'Math Talk'/'Try This' prompts through the running text, plus two formal 'Figure it Out' blocks. Every answer here is checked against the book's own printed answer key at the end of the chapter.. Questions are referenced from the NCERT textbook for identification.
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