Amount of molecule at equilibrium A cell produces molecule A at 10 units per second and the molecule has a half life of 15 seconds. How many units of molecule A are inside the cell at equilibrium? Showing your work would be phenomenal. Geez, a bit hostile, are we? Regardless, **Here's how I went about doing it. Pardon my terrible formatting skills:** P is the concentration of molecule A. The half-life formula is 2^-(t/15). At the moment of equilibrium: P = 10 * dt + 2^-(dt/15) * P Moving things around, I get P = 10 * dt / 1-2^-(dt/15) I attempt to take lim dt->0 using l'hopital's: 10 / (1/15 * ln(2)) = 150/ln(2) molecules of A at equilibrium. However, the correct answer should be 300 units of molecule A at equilibrium. Any suggestions?

I answered a very similar question from you here. Using the formulas on Wikipedia we can find that the mean lifetime $\tau$ is $T_{1/2}/\ln(2) = 15/\ln(2) \approx 21.6$. Using Little's Law the steady state concentration is (arrival rate) $\times$ (mean lifetime) $\approx 216$.

You could also use the decay rate $\lambda=\ln(2)/T_{1/2}$ and substitute it into the differential equation given in the previous answer.

This is the same as the answer that you got to by a more complicated route. I don't see how we can get to a steady-state concentration of 300 unless you've misstated the question or left something out.