# Exponential practice exam question

Okay bear with me, this is one of those cumulative questions

The amount of a certain type of drug in the bloodstream t hours after it has been taken is given by the formula:

$$x = D{e^{ - {1 \over 8}t}}$$

where x is the amount of the drug in the bloodstream in milligrams and D is the dose given in milligrams.

A dose of 10 mg of the drug is given.

(a) Find the amount of the drug in the bloodstream 5 hours after the dose is given.

\eqalign{ & x = D{e^{ - {1 \over 8}t}} \cr & x = 10{e^{ - {1 \over 8}\left( 5 \right)}} \cr & x = 5.353 \cr}

(b) A second dose of 10 mg is given after 5 hours. Show that the amount of the drug in the bloodstream 1 hour after the second dose is 13.549 mg to 3 decimal places.

So: \eqalign{ & 10{e^{ - {1 \over 8}\left( 6 \right)}} + 10{e^{ - {1 \over 8}\left( 1 \right)}} \cr & = 10{e^{ - {6 \over 8}}} + 10{e^{ - {1 \over 8}}} \cr & = 13.549 \cr}

Okay this is the part I'm stuck on:

(c) No more doses of the drug are given. At time T hours after the second dose is given, the amount of the drug in the bloodstream is 3 mg. Find the value of T.

How do I go about solving this, I know x=3, but how about D? There is no dose in this case, and t=t, so what do I do? Form some sort of simultaneous equation?

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Since the two doses were spaced 5 hours apart, the amount of the drug still present in the patient's bloodstream is $\ x = 10 \cdot [ \ e^{-t/8} + e^{-(t+5)/8} \ ]$ mg. , with $\ t \$ being the number of hours from the second dose; we must introduce a "time delay" into one of the exponents.

This actually doesn't make the equation difficult to solve, since we can use properties of exponents to write

$$\ x = 10 \cdot [ \ e^{-T/8} + e^{-(T+5)/8} \ ] = 10 \cdot [ \ e^{-T/8} \ + \ e^{-T/8} \cdot e^{-5/8} \ ]$$

$$= 10 \cdot [ \ 1 \ + \ e^{-5/8} \ ] \ \cdot \ e^{-T/8} \ = \ 3 .$$

The quantity in brackets is just a constant and the variable $\ T \$ that we wish to solve for only appears once.

Note -- we can make a quick estimate of the value of $\ T \$ to see if the calculated result is in the right neighborhood. The amount of drug present due to the second dose will dominate the amount remaining from the first, so we can consider the behavior of that dose. The "biological half-life" of the drug is given by $\ \tau = \frac{\ln 2}{1/8} \approx 5.5$ hours. The amount of drug present at $\ t = 0 \$ is about 13 mg. and 3 mg. is about one-quarter of this, so somewhat over two "half-lives" will be required to have the amount of drug fall to 3 mg. Thus, $\ T \$ should be somewhat larger than 11 hours (in fact, it comes out close to 13 hours).

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Thank you for the great answer, very appreciated. – seeker Jun 12 '13 at 20:22