With preconditions a > 1 and a < b (no overflow): int x = 0; int p = 1; while (p < b) { p = p * a; x = x + 1; } When the loop ends, x in terms of a and b is

1. Aaᵇ
2. Bbᵃ
3. C⌊log_a b⌋ (floor)
4. D⌈log_a b⌉ (ceiling)

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Correct answer: D. ⌈log_a b⌉ (ceiling)

Explanation

p takes the values a, a², … and stops at the first power with aˣ ≥ b, so x = ⌈log_a b⌉.