NCERT Solutions Ganita Prakash (Part 1) Chapter 2 –232.2 Exponential Notation and Operations — Figure it Out

Book page 22 Updated on2026-09-05

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

Answer

Count how many times each factor is repeated; that count becomes its exponent.

	Expanded form	Exponential form
(i)	6 × 6 × 6 × 6	6⁴ ( = 1296)
(ii)	y × y	y²
(iii)	b × b × b × b	b⁴
(iv)	5 × 5 × 7 × 7 × 7	5² × 7³ ( = 8575)
(v)	2 × 2 × a × a	2²a² ( = 4a²)
(vi)	a × a × a × c × c × c × c × d	a³c⁴d

Why it happens: different bases keep their own exponents; you may not add them. In (iv) there are five factors altogether, but they are not all the same number, so the answer is 5² × 7³ and not 5⁵ or 35⁵. In (vi), d appears once, so its exponent is 1 — and d¹ is simply written as d.

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

Answer

(i) 648 = 2 × 324 = 2 × 2 × 162 = 2 × 2 × 2 × 81
= 2³ × 3⁴ (8 × 81 = 648 ✓)

(ii) 405 = 5 × 81
= 3⁴ × 5 (81 × 5 = 405 ✓)

(iii) 540 = 2 × 270 = 2 × 2 × 135 = 4 × 27 × 5
= 2² × 3³ × 5 (4 × 27 × 5 = 540 ✓)

(iv) 3600 = 36 × 100 = (2² × 3²) × (2² × 5²)
= 2⁴ × 3² × 5² (16 × 9 × 25 = 3600 ✓)

Why it happens: every number has exactly one prime factorisation, so the exponential form is a fingerprint. Notice (iv): all three exponents are even, which is another way of saying 3600 is a perfect square — indeed 60², since 60 = 2² × 3 × 5.

Tip: break the number into two convenient factors first (3600 = 36 × 100), factorise each, then collect equal bases using n^a × n^b = n^a+b. It is much faster than dividing by 2 eleven times.

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

Answer

(i) 2 × 10³ = 2 × 1000 = 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) × 10,00,000 = –3,20,00,000

Why it happens: in (iv) both exponents are even, so both minus signs disappear and the answer is positive. In (vi) the exponents are 5 (odd) and 6 (even): the first factor stays negative, the second turns positive, and a negative times a positive is negative. You can also see it as (–3)² × (–5)² = (–3 × –5)² = 15² = 225, using m^a × n^a = (mn)^a.

Check it yourself: –3,20,00,000 is –3.2 × 10⁷ in the standard form you meet later in this chapter.

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