Dosage and half-life Problem

[JaysFanatic]

John finds an injured 30kg wolf. Thinking back to school, he remembered that the wolf would be anesthized if its bloodstream contained at least 45mg of sodium pentobarbital per kg. of weight. He also remembered that the half-life of the drug is 5 hours in a wolf or dog. He needs to anesthize the wolf for an 1.5 hours. What initial dose should he administer?

I found that the half-life of the drug was - ln(2)/8 and multiplied the weight of the wolf 30 kg by the amouny of the drug 45 mg to come up with 1350mg of the drug is at least needed.

I used the equation 1350= initial amount times e^(1.5)(-ln2/8) and got 1662.045 mg.
Is this right?

 

[tkhunny]

There is something missing from this problem. What is the lethal dosage? I hope it's less than 55 mg/kg!!

In any case, that looks fine. Why do you doubt?

 

[JaysFanatic]

I just wasn't sure whether the initial dosage should be higher or lower than the amount needed.

 

[lookagain]

. He also remembered that the half-life of the drug is 5 hours in a wolf or dog. He needs to anesthize the wolf for an 1.5 hours. What initial dose should he administer?

I found that the half-life of the drug was - ln(2)/8 and multiplied the weight of the wolf 30 kg by the amouny of the drug 45 mg to come up with 1350mg of the drug is at least needed.

I used the equation 1350= initial amount times e^\(\displaystyle [\)(1.5)(-ln2/8)\(\displaystyle ]\) and got 1662.045 mg.

\(\displaystyle > \ > \ >\)Regardless of the correctness of the factors in the exponent,
they have to have grouping symbols around them for the order of operations. \(\displaystyle < \ < \ <\)

In general. the k value is equal to -ln(2)/(half-life).


So k = [-ln(2)]/5, instead of [-ln(2)]/8.


\(\displaystyle Let \ Q_0 \ = \ initial \ amount \ of \ the \ dose \ in \ milligrams.\)


\(\displaystyle 1350 \ = \ Q_0*e^{\{([-ln(2)]/8)(1.5)\}}\)


\(\displaystyle Edit: \ \ \ Self \ correction\)


\(\displaystyle There \ is \ no \ problem \ with \ my \ formula.\)

I forgot the factor of 1.5 in the computation in the product in the exponent,
even though I had typed it in.


\(\displaystyle 1350 \ \approx \ Q_0(0.812252396)\)


\(\displaystyle Q_0 \ \approx \ 1662.045\)


\(\displaystyle The \ initial \ amount \ would \ be \ approximately \ 1662.045 \ mg.\)

 

[tkhunny]

That's why I never liked the traditional 'e' version of this problem. It's a half-life. Why not use 1/2, rather than e?

\(\displaystyle 1350 = Q_{0}\left\frac{1}{2}\right ^{\frac{1.5}{5}}\)

Q_{0} = 1662.045