Figure it Out — Exponents and Prime Factorisation — Power Play

1. 13 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.2), Q1

   Express the following in exponential form: (i) 6 × 6 × 6 × 6 (ii) y × y (iii) b × b × b × b (iv) 5 × 5 × 7 × 7 × 7 (v) 2 × 2 × a × a (vi) a × a × a × c × c × c × c × d

   Hint. Count how many times each distinct factor repeats.

   Step 1 — Count repeats of each base. (i) 6 repeats 4 times. (ii) y repeats 2 times. (iii) b repeats 4 times. (iv) 5 repeats twice, 7 repeats three times. (v) 2 repeats twice, a repeats twice. (vi) a repeats three times, c repeats four times, d repeats once, since each distinct letter is counted separately regardless of the order it appears in.

   ✦ Answer: (i) 6⁴ (ii) y² (iii) b⁴ (iv) 5² × 7³ (v) 2² × a² (vi) a³ × c⁴ × d.

   Where students slip. Miscounting repeated factors when several different letters/numbers are mixed together, as in (vi) — count each distinct base separately before combining.

2. 23 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.2), Q2

   Express each of the following as a product of powers of their prime factors in exponential form: (i) 648 (ii) 405 (iii) 540 (iv) 3600

   Hint. Prime-factorise each number completely, then group repeated primes into powers.

   Step 1 — Factorise 648. 648 = 2³ × 3⁴ (648 = 8 × 81).

   Step 2 — Factorise 405. 405 = 3⁴ × 5 (405 = 81 × 5).

   Step 3 — Factorise 540. 540 = 2² × 3³ × 5 (540 = 4 × 27 × 5).

   Step 4 — Factorise 3600. 3600 = 2⁴ × 3² × 5² (3600 = 16 × 9 × 25), since repeated division by each prime in turn accounts for every factor exactly once.

   ✦ Answer: (i) 2³ × 3⁴ (ii) 3⁴ × 5 (iii) 2² × 3³ × 5 (iv) 2⁴ × 3² × 5².

   Where students slip. Stopping the factorisation early (e.g. leaving a factor of 9 unsplit as 3×3) — keep dividing until only primes remain before grouping into powers.

3. 33 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.2), Q3

   Write the numerical value of each of the following: (i) 2 × 10³ (ii) 7² × 2³ (iii) 3 × 4⁴ (iv) (−3)² × (−5)² (v) 3² × 10⁴ (vi) (−2)⁵ × (−10)⁶

   Hint. Evaluate each power first, then multiply.

   Step 1 — (i) 2 × 10³ = 2 × 1000 = 2000.

   Step 2 — (ii) 7² × 2³ = 49 × 8 = 392.

   Step 3 — (iii) 3 × 4⁴ = 3 × 256 = 768.

   Step 4 — (iv) (−3)² × (−5)² = 9 × 25 = 225.

   Step 5 — (v) 3² × 10⁴ = 9 × 10000 = 90000.

   Step 6 — (vi) (−2)⁵ × (−10)⁶ = (−32) × (1000000) = −32000000, since an odd power of a negative number stays negative while an even power turns positive.

   ✦ Answer: (i) 2000 (ii) 392 (iii) 768 (iv) 225 (v) 90000 (vi) −32000000.

   Where students slip. Losing track of the sign in (vi) — (−2)⁵ is negative (odd power) while (−10)⁶ is positive (even power), so their product is negative.

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.