With t1/2 = 4 h and dosing interval τ = 12 h, the accumulation factor R is approximately:

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Multiple Choice

With t1/2 = 4 h and dosing interval τ = 12 h, the accumulation factor R is approximately:

Explanation:
Intermittent dosing with first-order elimination leads to a predictable accumulation described by R = 1 / (1 − e^(−kτ)), where k is the elimination rate constant (k = ln 2 / t1/2). With t1/2 = 4 h, k = 0.693 / 4 ≈ 0.1733 h⁻¹. For a dosing interval τ = 12 h, e^(−kτ) = e^(−0.1733 × 12) ≈ e^(−2.08) ≈ 0.125. Therefore R = 1 / (1 − 0.125) = 1 / 0.875 ≈ 1.14. This means at steady state the peak (and overall levels) are about 14% higher than after a single dose, reflecting modest accumulation because the interval is three half-lives.

Intermittent dosing with first-order elimination leads to a predictable accumulation described by R = 1 / (1 − e^(−kτ)), where k is the elimination rate constant (k = ln 2 / t1/2). With t1/2 = 4 h, k = 0.693 / 4 ≈ 0.1733 h⁻¹. For a dosing interval τ = 12 h, e^(−kτ) = e^(−0.1733 × 12) ≈ e^(−2.08) ≈ 0.125. Therefore R = 1 / (1 − 0.125) = 1 / 0.875 ≈ 1.14. This means at steady state the peak (and overall levels) are about 14% higher than after a single dose, reflecting modest accumulation because the interval is three half-lives.

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