After taking a tablet, a patient has 10 units/ml in a sample of blood taken soon after, and this decreased to 6 units/ml 9 hours later.
- What is the half-life of the tablet?
- How long will it take decrease to 20% of its original value?
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jaykirby
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Assuming exponential decay, you model the amount of drug $N$ in the bloodstream by $N=N_{0}e^{-kt}$. You can say that $N_{0}=10$, and then you can find $k$ by solving the equation $6=10e^{-9k}$ for it.
Once you have $k$, you can find the half-life, $t_{1/2}$, by solving $5=10e^{-kt_{1/2}}$. You can solve #2 in a similar fashion.
Adrian Keister
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Can I ask, what value did you get for k? Could someone walk through this step-by-step? – jaykirby Jul 19 '13 at 14:29
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Let's see: $0.6=e^{-9k}\implies \ln(0.6)=-9k\implies k=-\ln(0.6)/9\approx 0.0568$. – Adrian Keister Jul 19 '13 at 14:32
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And 28.24 hours for part 2? – jaykirby Jul 19 '13 at 14:33
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1If you show your work, I can comment on it. Your answer is in the right ball-park. – Adrian Keister Jul 19 '13 at 14:37
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1Sure. ln(0.20)= -0.057t. t= -ln(0.20)/0.057 = 28.24 – jaykirby Jul 19 '13 at 14:57
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1I'd probably keep one more significant digit of $k$. I get $28.34$ hours. – Adrian Keister Jul 19 '13 at 15:07
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So for the half-life of the tablet, I got 12.16 hours? did you find that too? – jaykirby Jul 20 '13 at 03:14
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Well, I'd have $0.5=e^{-0.0568t_{1/2}}$, so $\ln(0.5)=-0.0568t_{1/2}$, and therefore $$t_{1/2}=\frac{\ln(0.5)}{-0.0568}=12.2.$$ – Adrian Keister Jul 20 '13 at 11:50