Let n be any natural number such that 5ⁿ⁻¹

Question

Let n be any natural number such that 5ⁿ⁻¹<3ⁿ⁺¹. Then, the least integer value of m that satisfies 3ⁿ⁺¹<2ⁿ⁺ᵐ for each such n, is:

UPSC Quizroom · Accepted Answer

Solution: Step 1: Solve 5^(n−1) < 3^(n+1) Take logarithms: (n − 1) ln5 < (n + 1) ln3 → n ln5 − ln5 < n ln3 + ln3 → n(ln5 − ln3) < ln15 ln5 − ln3 ≈ 0.51083, ln15 ≈ 2.70805 → n < 2.70805 / 0.51083 ≈ 5.3 So n = 1, 2, 3, 4, 5 Step 2: Solve 3^(n+1) < 2^(n+m) Take logarithms: (n + 1) ln3 < (n + m) ln2 → m ln2 > n(ln3 − ln2) + ln3 → m > [n(ln3 − ln2) + ln3] / ln2 ln3 − ln2 ≈ 0.40546, ln3 / ln2 ≈ 1.585 Compute for n = 1..5: n=1 → m > 2.17 n=2 → m > 2.76 n=3 → m > 3.34 n=4 → m > 3.92 n=5 → m > 4.51 Maximum required m ≈ 4.51 → least integer m = 5 Answer: 5

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Let n be any natural number such that 5ⁿ⁻¹<3ⁿ⁺¹. Then, the least integer value of m that satisfies 3ⁿ⁺¹<2ⁿ⁺ᵐ for each such n, is:

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Let n be any natural number such that 5ⁿ⁻¹<3ⁿ⁺¹. Then, the least integer value of m that satisfies 3ⁿ⁺¹<2ⁿ⁺ᵐ for each such n, is:

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