How many half-lives are typically needed to reach approximately 95% steady-state for a first-order drug?

• A. 5
• B. 3
• C. 7
• D. 2

Answer: (A)

In first-order kinetics, the body eliminates a drug at a rate proportional to its concentration, so concentrations approach a steady value exponentially. A handy way to think about reaching steady state with repeated dosing is that after each half-life, the remaining gap to steady state halves. After n half-lives, the fraction of steady state achieved is 1 − (1/2)^n.

Doing the math for five half-lives gives 1 − (1/2)^5 = 1 − 1/32 ≈ 0.96875, about 97% of steady state—roughly the 95% mark clinicians use as “near steady state.” That’s why five half-lives is the typical answer.

Notes on the other options: fewer half-lives (like three or two) leave larger gaps to steady state (about 87.5% and 75%, respectively), while seven half-lives would exceed 95% by a comfortable margin, but the common rule-of-thumb for “approximately 95%” is five half-lives.