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Let m and n be positive real numbers such that m^(log₃ 5) = 9 and n^(log₅ 7) = 25. Also (log₃ m)*(log₅ n) = log₈ a, where b is a prime number. Match List I with List II. List I: (P) m^((log₃ 5)²) + n^((log₅ 7)²) (Q) value of a (R) value of b (S) antilog_(sqrt(b))(log_b 2401) List II: (1) 81 (2) 7 (3) 49 (4) 74

1. P -> 4; Q -> 1; R -> 2; S -> 3
2. P -> 4; Q -> 2; R -> 3; S -> 1
3. P -> 3; Q -> 1; R -> 2; S -> 4
4. P -> 1; Q -> 3; R -> 2; S -> 4

Correct answer: P -> 4; Q -> 1; R -> 2; S -> 3

Solution

P: m^((log₃ 5)²) = [m^(log₃ 5)]^(log₃ 5) = 9^(log₃ 5) = (3²)^(log₃ 5) = 3^(2*log₃ 5) = 5² = 25. Similarly n^((log₅ 7)²) = 25^(log₅ 7) = 5^(2*log₅ 7) = 7² = 49. P = 74 -> (4). For Q: log₃ m = 2/log₃ 5 (from first condition), log₅ n = 2/log₅ 7 (from second). Product = 4/[(log₃ 5)(log₅ 7)] = 4/log₃ 7. Matching with log₈(81): log₈ 81 = log₈(3⁴) = 4*(log₈ 3). And 4/log₃ 7 = 4*log₇ 3. These are equal when log₈ 3 = log₇ 3, which requires 8=7, untrue. The problem intends a=81 from the list matching, Q->(1). R=b=7 (prime) -> (2). S = antilog_(sqrt 7)(log₇ 2401). log₇ 2401 = log₇(7⁴)=4. antilog_(sqrt 7)(4) = (sqrt 7)⁴ = 7² = 49 -> (3).

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