NCERT Solutions Ganita Prakash (Part 1) Chapter 2 .3 The Other Side of Powers — In-text Questions

Book page 292 Updated on2026-09-05

Q1.

Can we write 10³ = 1/10⁻³ ?

Answer

Yes.

1 ÷ 10^–3 = 1 ÷ (1/10³)
= 1 × 10³ (dividing by a fraction means multiplying by its reciprocal)
= 10³

The same argument gives 7² = 1/7^–2 and 4^a = 1/4^–a. In general

n^–a = 1/n^a and n^a = 1/n^–a, where n ≠ 0

Why it happens: a negative exponent means "reciprocal", and taking the reciprocal twice returns you to where you started. So the minus sign in an exponent can always be moved by shifting the power across the fraction bar — from the denominator to the numerator, or back. This is the single most useful habit in the whole topic: a power crossing the bar changes the sign of its exponent.

Q2.

We had required a and b to be counting numbers. Can a and b be any integers? Will the generalised forms still hold true?

Answer

Yes — once n⁰ = 1 and n^–a = 1/n^a are defined as they are, all three rules hold for every integer a and b (with n ≠ 0).

Check the product rule where one exponent is negative:

2^–4 × 2⁷ = (1/2⁴) × 2⁷ = 2⁷/2⁴ = 2³
and the rule predicts 2^–4+7 = 2³ ✓

3^–2 × 3^–5 = (1/3²) × (1/3⁵) = 1/3⁷ = 3^–7
and the rule predicts 3^–2–5 = 3^–7 ✓

(13^–2)^–3 = 1/(13^–2)³ = 1/13^–6 = 13⁶
and the rule predicts 13^{(–2)×(–3)} = 13⁶ ✓

Why it happens: the definitions were not chosen at random — they were chosen so that the rules would survive. The chapter finds n⁰ by insisting that 2⁴ ÷ 2⁴ = 2^4–4 must still be true, and finds n^–a by insisting that 2⁴ ÷ 2⁵ = 2^4–5 must still be true. Extending a rule and then defining the new symbols so the rule keeps working is a move mathematics uses constantly.

Q3.

Write equivalent forms of the following. (i) 2⁻⁴ (ii) 10⁻⁵ (iii) (– 7)⁻² (iv) (– 5)⁻³ (v) 10⁻¹⁰⁰

Answer

	Given	Equivalent form	Value
(i)	2^–4	1/2⁴	1/16 = 0.0625
(ii)	10^–5	1/10⁵	1/1,00,000 = 0.00001
(iii)	(–7)^–2	1/(–7)²	1/49 (positive)
(iv)	(–5)^–3	1/(–5)³	–1/125 (negative)
(v)	10^–100	1/10¹⁰⁰	a decimal point, 99 zeros, then 1

Why it happens: the minus in the exponent flips the number over; it does not make the number negative. That is why (iii) is positive — its sign is decided by the even exponent 2, not by the minus in front of it — while (iv) is negative because 3 is odd. A negative exponent always produces a number between 0 and 1 when the base is bigger than 1.

Did you know? 10¹⁰⁰ is called a googol, so 10^–100 is one googolth — smaller than the size of an atom compared with the size of the universe.

Q4.

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 × 8ᵠ

Answer

Same base each time, so add the exponents.

(i) 2^–4 × 2⁷ = 2^–4+7 = 2³ ( = 8)
(ii) 3² × 3^–5 × 3⁶ = 3^2–5+6 = 3³ ( = 27)
(iii) p³ × p^–10 = p^3–10 = p^–7 ( = 1/p⁷)
(v) 8^p × 8^q = 8^p+q

Part (iv) needs one extra step, because the bases 2 and –4 are different. Rewrite –4 as a power of 2, remembering that the exponent is even so the sign disappears:

(iv) (–4)^–2 = 1/(–4)² = 1/16 = 2^–4
2⁴ × (–4)^–2 = 2⁴ × 2^–4 = 2^4–4 = 2⁰ = 1

Why it happens: (iii) is worth a second look — a negative answer for the exponent is perfectly acceptable. p^–7 is simply 1/p⁷, a number smaller than 1 when p > 1. Nothing has gone wrong; the exponent has just crossed zero, exactly as the power line on the next page shows.

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