Power Play

Why powers matter

Suppose a sheet of paper is 0.001 cm thick. Every ideal fold doubles its thickness. After n folds, thickness = 0.001 × 2ⁿ cm. Ten folds give 1.024 cm, 20 give about 10.49 m, 30 about 10.74 km, and 46 about 703,687 km. Real paper cannot practically be folded this many times: the model illustrates mathematical exponential growth, not a feasible experiment.

2.1 Experiencing the Power Play

Exponential growth multiplies by a constant factor; linear growth adds a constant amount. A 20 cm ladder step increases height by 20 cm each time, whereas ideal folding doubles thickness. A pond whose lotus cover doubles daily and is full on day 30 is half-covered on day 29. Compare sequences: linear 2,4,6,8 and exponential 2,4,8,16. When modelling, identify assumptions and units.

2.2 Exponential notation and operations

A power aⁿ has base a and exponent n. For positive integer n, aⁿ means n repeated factors of a. Examples: 5⁴=625, (−4)³=−64, (−2)⁴=16, and 0⁵=0. Prime factorisation: 32400=2⁴×3⁴×5². A product of powers of the same nonzero base adds exponents: aᵐ×aⁿ=aᵐ⁺ⁿ. Power of a power multiplies exponents: (aᵐ)ⁿ=aᵐⁿ. For equal exponents, aⁿbⁿ=(ab)ⁿ. Dividing equal-base powers subtracts exponents: aᵐ÷aⁿ=aᵐ⁻ⁿ, a≠0. The number of possible 5-position numeric codes with leading zero allowed is 10⁵=100000. Six alphabetic positions with 26 choices each give 26⁶ codes.

2.3 The Other Side of Powers

For nonzero a, a⁰=1 and a⁻ⁿ=1/aⁿ. Examples: 2⁻³=1/8; (−5)⁻³=−1/125; 10⁻⁵=0.00001. The expression 0⁰ is not defined by these rules; 0 raised to a negative power is undefined. A power line for base 4 moves right by multiplying by 4 and left by dividing by 4: 4⁻²=1/16, 4⁻¹=1/4, 4⁰=1, 4¹=4, 4²=16.

2.4 Powers of 10 and scientific notation

Place values can be represented using integer powers of ten: 561.903=5×10²+6×10¹+1×10⁰+9×10⁻¹+0×10⁻²+3×10⁻³. Scientific notation writes nonzero positive numbers as c×10ⁿ where 1≤c<10. Examples: 5900=5.9×10³; 59853=5.9853×10⁴; 0.00072=7.2×10⁻⁴; 8000000=8×10⁶. For negative numbers, use −c×10ⁿ. The exponent records order of magnitude; the coefficient preserves precision.

2.5 Did You Ever Wonder? Estimation and modelling

Large-number problems need explicit assumptions. Example: 45 kg of jaggery at ₹70/kg costs ₹3150. A 400 km walk at 5 km/h takes 80 walking hours, not counting rest. If a person sleeps eight hours daily, one year of sleep is roughly 365×8×3600≈1.05×10⁷ seconds. Approximation is not the same as an exact measurement; explain your assumed quantities and avoid presenting estimates as verified current statistics.

2.5 A Pinch of History

The textbook introduces historical Indian names for large powers of ten and references the Lalitavistara, Mahaviracharya, and Jain and Buddhist mathematical traditions. One lakh=10⁵, one crore=10⁷, one arab=10⁹ and one kharab=10¹¹. International terms include million=10⁶, billion=10⁹ and trillion=10¹². A googol is 10¹⁰⁰. Historical anecdotes and currency claims are textbook context, not necessarily current facts.

Worked examples and exam pitfalls

Example 1: 2⁻⁴×2⁷=2³=8. Example 2: (13⁻²)⁻³=13⁶. Example 3: 2⁴×(−4)⁻²=16×1/16=1. Example 4: 8ᵖ×8ᑫ=8ᵖ⁺ᑫ. Avoid adding bases in a²×b²: it equals (ab)², not (a+b)². 2³+2³=16, not 2⁶. Parentheses matter: −3²=−9 but (−3)²=9.

Formula sheet and quick revision

aᵐaⁿ=aᵐ⁺ⁿ; aᵐ/aⁿ=aᵐ⁻ⁿ (a≠0); (aᵐ)ⁿ=aᵐⁿ; aⁿbⁿ=(ab)ⁿ; aⁿ/bⁿ=(a/b)ⁿ (b≠0); a⁰=1 (a≠0); a⁻ⁿ=1/aⁿ (a≠0); ideal fold thickness=t₀×2ⁿ; scientific notation=c×10ⁿ with 1≤c<10 for positive c. Check that the same-base laws only combine like bases and that negative exponents do not mean negative values.

Original exam practice, MCQs and HOTS

MCQ: 3⁴×3² equals (A)3⁶ (B)9⁶ (C)3⁸. Answer A, since add exponents. MCQ: 10⁻³ equals (A)1000 (B)0.001 (C)−1000. Answer B. HOTS: Find x if 2ˣ=128. Since 128=2⁷, x=7. Case study: A bacterial culture doubles every hour from 200 cells. After five hours it contains 200×2⁵=6400 cells, assuming uninterrupted doubling. Competency: A 4-digit PIN with leading zeros has 10⁴=10000 possibilities. Explain why 0000 is included.

Self-assessment and flashcards

Check yourself: (1) 5⁰=1; (2) 4⁻²=1/16; (3) 2⁷×2³=2¹⁰; (4) 7.2×10⁵=720000; (5) 10⁻⁴=0.0001. Flashcards: Base—repeated factor; exponent—number of factors; negative exponent—reciprocal; scientific notation—coefficient times power of ten; exponential growth—constant multiplication factor.

Source and coverage statement

Source: NCERT Ganita Prakash Grade 8 Part I Chapter 2, reprint 2026–27, pages 19–46. Exact textbook wording and figures remain in the official PDF at https://ncert.nic.in/textbook/pdf/hegp102.pdf. The 3 Figure it Out questions on pages 22–23 and 14 on pages 44–45 are addressed below; exploratory modelling tasks have illustrative answers where assumptions are required. Additional practice is original, not labelled as NCERT.

NCERT Figure it Out — pages 22–23 (Questions 1–3)

Each item provides a complete question summary, worked reasoning and a final answer. Consult the official textbook for the original wording and diagrams.

1. Question 1. Express repeated products in exponential form

   Step-by-step solution: (i) 6×6×6×6=6⁴; (ii) y×y=y²; (iii) b×b×b×b=b⁴; (iv) 5×5×7×7×7=5²×7³; (v) 2×2×a×a=2²a²; (vi) a×a×a×c×c×c×c×d=a³c⁴d. Count identical factors separately.

   Final answer: (i) 6⁴; (ii) y²; (iii) b⁴; (iv) 5²×7³; (v) 2²a²; (vi) a³c⁴d.

2. Question 2. Express 648, 405, 540 and 3600 as products of prime powers

   Step-by-step solution: Divide each number repeatedly by its prime factors: (i) 648=8×81=2³×3⁴; (ii) 405=81×5=3⁴×5; (iii) 540=4×27×5=2²×3³×5; (iv) 3600=16×9×25=2⁴×3²×5².

   Final answer: (i) 2³×3⁴; (ii) 3⁴×5; (iii) 2²×3³×5; (iv) 2⁴×3²×5².

3. Question 3. Evaluate the six numerical expressions

   Step-by-step solution: (i) 2×10³=2000; (ii) 7²×2³=49×8=392; (iii) 3×4⁴=3×256=768; (iv) (−3)²×(−5)²=9×25=225; (v) 3²×10⁴=9×10000=90000; (vi) (−2)⁵×(−10)⁶=−32×1000000=−32000000.

   Final answer: 2000; 392; 768; 225; 90000; −32000000.

NCERT Figure it Out — pages 44–45 (Questions 1–14)

Each item provides a complete question summary, worked reasoning and a final answer. Consult the official textbook for the original wording and diagrams.

1. Question 1. Find the units digit of 2²²⁴ ÷ 4³²

   Step-by-step solution: Write 4³²=(2²)³²=2⁶⁴. Thus the expression equals 2¹⁶⁰. The units digits of powers of 2 repeat 2,4,8,6 every four exponents. Since 160 is divisible by 4, the last digit is 6.

   Final answer: 6.

2. Question 2. A new container with five bottles arrives daily. How many bottles after 40 days?

   Step-by-step solution: Assuming no container initially and one new container each day, 40 containers × 5 bottles = 200 bottles. This is linear growth, not exponential growth.

   Final answer: 200 bottles.

3. Question 3. Write 64³, 192⁸ and 32⁻⁵ as products of two or more powers in three ways each

   Step-by-step solution: (i) 64³=(2³×2³)³=2⁹×2⁹=(4×16)³=4³×16³=(8×8)³=8³×8³. (ii) 192⁸=(64×3)⁸=64⁸×3⁸=(2⁵×6)⁸=2⁴⁰×6⁸=(12×16)⁸=12⁸×16⁸. (iii) 32⁻⁵=(2×16)⁻⁵=2⁻⁵×16⁻⁵=(4×8)⁻⁵=4⁻⁵×8⁻⁵=(2²×2³)⁻⁵=2⁻¹⁰×2⁻¹⁵.

   Final answer: Three valid factorisations for each expression, as shown.

4. Question 4. Classify five claims as always, sometimes or never true

   Step-by-step solution: (i) A cube is also a square only when its prime exponents are multiples of 6: sometimes (64 yes, 8 no). (ii) Every integer fourth power is a square because n⁴=(n²)²: always. (iii) For nonzero integer n, n⁵/n³=n² is an integer: always for nonzero integers; zero requires care because division by 0 is undefined. (iv) Product a³b³=(ab)³: always. (v) For prime q, q⁴⁶ has exponent 46, not divisible by 4 or 6: never.

   Final answer: (i) Sometimes; (ii) Always; (iii) Always for nonzero integers; (iv) Always; (v) Never.

5. Question 5. Simplify the five exponential expressions

   Step-by-step solution: (i) 10⁻²×10⁻⁵=10⁻⁷. (ii) 5⁷÷5⁴=5³. (iii) 9⁻⁷÷9⁴=9⁻¹¹. (iv) (13⁻²)⁻³=13⁶. (v) m⁵n¹²(mn)⁹=m⁵n¹²m⁹n⁹=m¹⁴n²¹.

   Final answer: 10⁻⁷; 5³; 9⁻¹¹; 13⁶; m¹⁴n²¹.

6. Question 6. Use 12²=144 to calculate four related squares

   Step-by-step solution: (i) (1.2)²=(12/10)²=144/100=1.44. (ii) (0.12)²=144/10000=0.0144. (iii) (0.012)²=144/1000000=0.000144. (iv) 120²=(12×10)²=144×100=14400.

   Final answer: 1.44; 0.0144; 0.000144; 14400.

7. Question 7. Circle equivalent expressions among five given powers

   Step-by-step solution: Compare 2⁴×3⁶, 6⁴×3², 6¹⁰, 18²×6², and 6²⁴. Convert 6⁴×3²=(2×3)⁴×3²=2⁴×3⁶. Also 18²×6²=(3×6)²×6²=3²×6⁴=2⁴×3⁶. The other expressions differ.

   Final answer: 2⁴×3⁶, 6⁴×3² and 18²×6².

8. Question 8. Identify the greater number in three comparisons

   Step-by-step solution: (i) 4³=64 and 3⁴=81, so 3⁴ is greater. (ii) 2⁸=256 and 8²=64, so 2⁸ is greater. (iii) 100²=10000 whereas 2¹⁰⁰ is far larger, so 2¹⁰⁰ is greater.

   Final answer: (i) 3⁴; (ii) 2⁸; (iii) 2¹⁰⁰.

9. Question 9. How many digits are needed for 8.5 billion unique milk-packet IDs?

   Step-by-step solution: With digits 0–9 and leading zeros permitted, d positions give 10ᵈ codes. Eight positions give 100 million, too few; nine give one billion, still too few; ten give ten billion, enough.

   Final answer: 10 digits.

10. Question 10. Find numbers that are both squares and cubes and describe the pattern

    Step-by-step solution: An integer that is both a square and a cube has every prime exponent divisible by both 2 and 3, hence by 6. Examples: 64=2⁶, 729=3⁶ and 4096=4⁶.

    Final answer: Perfect sixth powers, n⁶.

11. Question 11. Count possible five-character alphanumeric locker codes

    Step-by-step solution: Each position can contain any of 10 digits or 26 English letters: 36 choices. With repetitions and leading zeros permitted, total=36⁵=60466176.

    Final answer: 60,466,176 codes.

12. Question 12. Combine approximate sheep and goat populations

    Step-by-step solution: The textbook gives approximately 10⁹ sheep and 10⁹ goats. Sum=10⁹+10⁹=2×10⁹. This is an approximate textbook modelling value, not a current census.

    Final answer: 2×10⁹.

13. Question 13. Estimate clothing items, honeybees, bacterial cells and eating time

    Step-by-step solution: (i) Using textbook world population ≈8.2×10⁹, 30 items/person gives 2.46×10¹¹ items. (ii) 10⁸ bee colonies ×5×10⁴ bees=5×10¹² bees. (iii) 3.8×10¹³ bacteria/person ×8.2×10⁹ people=3.116×10²³ bacterial cells. (iv) Illustrative assumptions: 70 years, 3 meals/day, 30 minutes/meal: 70×365×3×1800=137970000 seconds≈1.38×10⁸ seconds. Values are approximate and depend on assumptions.

    Final answer: (i) 2.46×10¹¹; (ii) 5×10¹²; (iii) 3.116×10²³; (iv) ≈1.38×10⁸ seconds (assumed).

14. Question 14. What date was one billion seconds before the calculation date?

    Step-by-step solution: One billion seconds equals 1000000000÷86400≈11574.07 days, or about 31.69 years. The exact calendar date depends on the chosen reference date and time zone. Do not report a fixed textbook date without specifying the reference instant.

    Final answer: Approximately 31 years and 8 months earlier; exact date is reference-dependent.

In-text textbook activities and worked responses

1. Paper-folding thickness table

   Formula t=0.001×2ⁿ cm. For n=21, t=2097.152 cm≈20.97 m; n=26 gives 67108.864 cm≈671.09 m; n=30 gives 1073741.824 cm≈10.74 km; n=40 gives 10995116.28 m≈10995 km; n=46 gives 703687.44 km. This is an ideal mathematical model.

2. Three daughters and diamonds

   With seven successive factors of three, total diamonds=3⁷=2187.

3. Lotus ponds

   A doubling pond full on day 30 is half-full on day 29. Four days doubling followed by four days tripling gives 2⁴×3⁴=(2×3)⁴=6⁴=1296 lotuses, assuming one initial lotus.

4. Combinations and passwords

   Seven dresses, two hats and three shoe pairs give 7×2×3=42 combinations. A five-position numeric lock gives 10⁵=100000 possibilities. A six-position A–Z lock gives 26⁶=308915776 possibilities.

5. Decimal expanded form

   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.

6. Scientific notation practice

   59853=5.9853×10⁴; 65950=6.595×10⁴; 3430000=3.43×10⁶; 70040000000=7.004×10¹⁰.

7. Modelling and estimates

   Use assumptions openly. For 400 km at 5 km/h, walking time=80 h; if walking 8 h/day, about 10 days, excluding extra rest. A billion seconds≈31.69 years. Such results are estimates, not universal facts.