SOLUTION: Hello E-Instructors :) Would someone kindly show me how to solve this problem correctly? Problem: The level of a prescription drug in the human body over time can be fou

Question 150694: Hello E-Instructors :)

Would someone kindly show me how to solve this problem correctly?

Problem: The level of a prescription drug in the human body over time can be found using the formula:
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.
(''''These marks are merely spacers and not intended to be part of the equation).

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?

So far I think that the formula may look like this:
2.5 = ___24________
'''''''''''''1-(1/2)^24/14

Then maybe this:
2.5 = ___24____
'''''''''''''0.3047534

2.5 = 69.057991

Then, I don't know what to do next or if what I've done is even close to being right. :P

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?

I'm completely in the dark about this one.

3. What is the difference between taking 2.5 milligram of Lorazepam
every 12 hours and taking 5 milligrams every 24 hours?

This one, I might subtract results from these two equations:

2.5 = ___12_________
''''''''''''1-(1/2)^12/14
from
5 = ___24_______
''''''''''1-(1/2)^24/14

Well, that would be after I have learned how to solve the first question
correctly. ;)

Thank you so much for sharing your valuable knowledge with a mathematically challenged damsel like myself.
Grateful,
J. ;)
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
1)
I assume that this is the formula.

L=3.59585 mg
.
2)

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=11.1618 mg
.
Ed

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