Half life question in matlab

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?

Answers (2)

Star Strider
Star Strider on 29 Jan 2015
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 Comments

Bradley Sanchez’ ‘Answer’ moved here...
So would I just assign h as a variable with the value of 7.9. Then do solve(-ln(2)/h), in matlab.
Well, only if you want to get the wrong answer!
‘Note that the decay constant is −ln(2)/H.’
The differential equation that describes single-compartment kinetics is:
dC(t)/dt = K*C(t)
that integrates to an exponential function, with K = -log(2)/H.
Half the initial concentration ‘C(0)’ will be eliminated in 7.9 hours. Since 20/50 is less than half, you already know that the concentration of 20mg is less than the half-life concentration of 25mg, so the time you will solve for will be greater than 7.9 hours.
Code up your equation, then use solve to find the solution the problem asks for.
If you have problems getting your code to work, post what you have here and we can help you get it running.
This is all I have I am still confused on what you would actually put in the solve equation? k= -log(2)/h p5=solve(
You have to put an equation of some sort in the argument to solve.
How would you calculate the clearance curve on paper with a calculator? How would you solve it for the time at which the concentration is 20mg? (That might be a good idea anyway, so you’ll know if your MATLAB code gives the correct result.) You have all the information you need.
The world will not end if your code doesn’t run the first time, or if it gives the wrong result. Please experiment with it.
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.
And if it with its Graph ?

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Roger Stafford
Roger Stafford on 29 Jan 2015
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.

2 Comments

Bob
Bob on 29 Jan 2015
Edited: Bob on 29 Jan 2015
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?
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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Asked:

Bob
on 29 Jan 2015

Commented:

on 10 May 2020

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