Half life question in matlab

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Bob
Bob 2015 年 1 月 29 日
コメント済み: noor iraq 2020 年 5 月 10 日
Metronidazole (used to treat infections) has a half life of H = 7.9 hours. Use solve to calculate [5 pt] the amount of time it takes for 50mg to decay to 20mg. Note that the decay constant is −ln(2)/H.
Can someone help me with the code that would be used to solve this type of problem?

回答 (2 件)

Star Strider
Star Strider 2015 年 1 月 29 日
See the documentation for the solve function. The question itself has all the information you need otherwise. Particularly note ‘−ln(2)/H’.
The question assumes single-compartment kinetics for metronidazole, an anti-protozoal agent that is also active against C. dificile and some others.
  6 件のコメント
Star Strider
Star Strider 2015 年 1 月 30 日
For the record, the solution to this problem is quite simply:
syms C(t)
H = 7.9;
K = -log(2)/H;
C(t) = exp(K*t);
t20s = solve(C == 20/50, t);
t20d = vpa(t20s, 3)
giving:
t20d =
10.4
So the concentration has reached the 20mg level at 10.4 hours.
noor iraq
noor iraq 2020 年 5 月 10 日
And if it with its Graph ?

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Roger Stafford
Roger Stafford 2015 年 1 月 29 日
Bradley, just ask yourself, "what power of 2 do I have to divide 50 by to get 20?". Then that power should be the ratio of the necessary decay time to 7.9 hours. It would be a very simple one-line matlab computation.
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Bob
Bob 2015 年 1 月 29 日
編集済み: Bob 2015 年 1 月 29 日
so that would be 20 because 20^2 then divide that by 20 which gives you 20. So I have this: solve(7.9*(-log(2)/(log(50/20)))). But I am getting an error because I don't have a variable. Would I replace 7.9 with x?
Roger Stafford
Roger Stafford 2015 年 1 月 30 日
If you discard that incorrect minus sign, you've already solved your problem without using 'solve'. They ought to give you extra credit for that.
If you are determined to use 'solve', then go back a step and write the equation you were trying to solve before you took the logarithm.

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