SOLUTION: Ok so I have an exponential and log functions activity and I am stuck on 1 problem. I did the first question but I am a little confused on the second one. Here is some backgr

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Ok so I have an exponential and log functions activity and I am stuck on 1 problem. I did the first question but I am a little confused on the second one. Here is some backgr      Log On


   



Question 898293: Ok so I have an exponential and log functions activity and I am stuck on 1 problem.
I did the first question but I am a little confused on the second one.
Here is some background info for the activity.

Exponential functions can be used to model the concentration of a drug in a patient's body. Suppose the concentration of Drug X in a patient's bloodstream is modeled by,
C (t) = C0 e - rt,
where C (t) represents the concentration at time t (in hours), C0 is the concentration of the drug in the blood immediately after injection, and r > 0 is a constant indicating the removal of the drug by the body through metabolism and/or excretion. The rate constant r has units of 1/time (1/hr). It is important to note that this model assumes that the blood concentration of the drug (C0 ) peaks immediately when the drug is injected.


1) suppose the drug has a removal rate constant of r = 0.080 1/hr and an initial concentration (Co) of 5.0 mg/L. Find the drug concentration after 4.0 hours using C(t) = Co e^-rt,.
I got C(4) = 5e ^ (-0.08)(4) = 3.6 mg/L
The second question asks
2) what percent of the drug is left in the body after this time? Use units of measure.
would it be the previous answer in percent form?
thanks so much!

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
C%28t%29 is in milligrams per liter. This is not PERCENT, but instead a measure of mass of the drug in 1 liter (of the patient's blood).

Question Number 2 then is asking for %28C%284%29%2FC%280%29%29%2A100;

In more specific values, %283.6%2F5.0%29%2A100 percent.