Exponents Test - 1

2³ = 2 × 2 × 2

= 8

3² = 3 × 3

= 9

⇒ 2³ + 3² = 8 + 9

= 17

(–2)³ × (–7)³ = {(–2) × (–2) × (–2)} × {(–7) × (–7) × (–7)}

= (–8) × (–343) = 2744

∴(–2)³ × (–7)³ = 2744

Thus, the relation given in alternative C is correct.

3² = 3 × 3 = 9

2³ = 2 × 2 × 2 = 8

∴ 3² > 2³

The given expressions 2⁶, 3⁴, 5³, and 11² can be expanded as

2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

3⁴ = 3 × 3 × 3 × 3 = 81

5³ = 5 × 5 × 5 = 125

11²= 11 × 11 = 121

It can be seen that 125 > 121 > 81 > 64.

∴5³ > 11² > 3⁴ > 2⁶

Thus, the given expressions 2⁶, 3⁴, 5³, and 11² can be arranged in a decreasing order as 5³ > 11² > 3⁴ > 2⁶.

The given equation is 5^m × 25² × 125^–1 = 1.

5^m × 25² × 125^–1 = 1

⇒ 5^m × (5²)² × (5³)^–1 = 1

⇒ 5^m × 5^{2 × 2} × 5^{3 × (–1)} = 1                           [(a^m)ⁿ = a^mn]

⇒ 5^m × 5⁴ × 5^–3 = 1

⇒ 5^{m + 4 – 3} = 1                                              [a^m × aⁿ = a^m + n]

⇒ 5^{m + 1} = 1

⇒ 5^{m + 1} = 5⁰                                                             [a⁰ = 1 (a ≠ 0)]

Since the bases are equal, their powers can be equated.

∴ m + 1 = 0

⇒ m = 0 – 1 = –1

Thus, the required value of m is –1.

The given number 51840 can be factorised as:

2	51840
2	25920
2	12960
2	6480
2	3240
2	1620
2	810
3	405
3	135
3	45
3	15
5	5
	1

∴ 51840 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 = 27 × 34 × 5

Comparing 2⁷ ×34× 5 with 2^a × 3^b × 5, we obtain a = 7 and b = 4

Thus, the respective values of a and b are 7 and 4.