SOLUTION: L=D/1-(1/2)^n/H where D is the amount taken every n hours and H is the drug’s half-life in hours. 1. If 2.5 milligrams of Lorazepam with a half-life of 14 hours is t

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: L=D/1-(1/2)^n/H where D is the amount taken every n hours and H is the drug’s half-life in hours. 1. If 2.5 milligrams of Lorazepam with a half-life of 14 hours is t      Log On

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Question 150680: L=D/1-(1/2)^n/H



where D is the amount taken every n hours and H is the drug’s half-life in hours.
1. If 2.5 milligrams of Lorazepam with a half-life of 14 hours is taken every 24 hours, then to what level does the drug build up over time?
2. If a doctor wants the level of Lorazepam to build up to a level of 5.58 milligrams in a patient taking 2.5 milligram doses, then how often should the doses be taken?
3. What is the difference between taking 2.5 milligrams of Lorazepam every 12 hours and taking 5 milligrams every 24 hours?

Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
1)
L=%28D%2F%281-%28%281%2F2%29%5E%28n%2FH%29%29%29%29 I assume that this is the formula.
L=%282.5%2F%281-%28%281%2F2%29%5E%2824%2F14%29%29%29%29
L=%282.5%2F%281-.304753%29%29
L=2.5%2F.695247
L=3.59585 mg
.
2)
L=%28D%2F%281-%28%281%2F2%29%5E%28n%2FH%29%29%29%29
5.58=%282.5%2F%281-%28%281%2F2%29%5E%28n%2F14%29%29%29%29
Let x=1-((1/2)^(n/14))
5.58=2.5/x
x=2.5/5.58
x=.448029
Let y=(1/2)^(n/14)
1-y=.448029
-y=.448029-1
y=1-.448029
y=.551971
Let z=n/14
.5^z=.551971
log[.5](.5^z)=log[.5](.551971)
z=log[.5](.551971)
log[.5](.551971)=log[10](.551971)/log[10](.5)=.857336
n/14=.857336
n=14*.857336
=12.0027 hrs
So 2.5 mg must be given every 12 hrs.
.
3)
L=%285%2F%281-%28%281%2F2%29%5E%2812%2F14%29%29%29%29
L=11.1618 mg
.
Ed