We test values:

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• Peter studies algorithm efficiency and compares two sorting methods. Method A runs in O(n²) time and takes 0.01n² seconds for n elements. Method B runs in O(n log n) time and takes 0.05n log₂(n) seconds. For what minimum value of n does Method B become faster than Method A?
• We need to solve: 0.05n log₂(n) < 0.01n².
• Divide both sides by n (n > 0): 0.05 log₂(n) < 0.01n → log₂(n) < 0.2n.
• n = 10: log₂(10) ≈ 3.32, 0.2×10 = 2 → 3.32 > 2 → A faster
• n = 20: log₂(20) ≈ 4.32, 0.2×20 = 4 → 4.32 > 4
• n = 25: log₂(25) ≈ 4.64, 0.2×25 = 5 → 4.64 < 5 → B faster

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