The quantity, Q
, of a drug in the blood stream begins with 300
 mg and decays to one-fifth its value over every 120
 minute period. Give the constants a
, b
, and T
 so that Q
 is represented by a function of the form Q=ab−t/T
, where t
 is the time in minutes after the drug is injected.

To represent the quantity \( Q \) of a drug in the bloodstream as a function of time \( t \) using the form \( Q = ab^{-t/T} \), we need to determine the constants \( a \), \( b \), and \( T \) given the initial conditions and the decay behavior of the drug.

### Given Information:
- Initial quantity of the drug: \( Q(0) = 300 \) mg.
- The drug decays to one-fifth of its value every 120 minutes.

### Function Form:
The function form given is:

\[
Q(t) = ab^{-t/T}
\]

where:
- \( a \) is the initial quantity of the drug.
- \( b \) is the base of the exponential decay.
- \( T \) is the time constant associated with the decay.

### Step-by-Step Solution:

1. **Determine \( a \):**

   The initial quantity of the drug in the bloodstream is 300 mg. Therefore, when \( t = 0 \):

   \[
   Q(0) = ab^{0} = a \times 1 = a
   \]

   So, \( a = 300 \).

2. **Determine \( b \) and \( T \):**

   The problem states that the drug decays to one-fifth of its value every 120 minutes. This means:

   \[
   Q(120) = \frac{1}{5} Q(0)
   \]

   Substituting \( Q(0) = 300 \) and the functional form:

   \[
   Q(120) = 300b^{-120/T} = \frac{1}{5} \times 300
   \]

   Simplifying this:

   \[
   300b^{-120/T} = 60
   \]

   Dividing both sides by 300:

   \[
   b^{-120/T} = \frac{60}{300} = \frac{1}{5}
   \]

   Taking the reciprocal to simplify:

   \[
   b^{120/T} = 5
   \]

3. **Solve for \( b \):**

   We know that \( b^{120/T} = 5 \), and since we want a base for our exponential decay:

   \[
   b = 5^{T/120}
   \]

   To maintain a consistent decay form, we can choose \( T = 120 \) because every 120 minutes, the drug reduces to one-fifth:

   \[
   b = 5^{120/120} = 5
   \]

4. **Final Values for \( a \), \( b \), and \( T \):**

   - \( a = 300 \) (initial quantity of the drug)
   - \( b = 5 \) (base of the exponential decay)
   - \( T = 120 \) (time constant for decay to one-fifth)

### Final Function:

The function that represents the quantity \( Q \) of the drug in the bloodstream over time \( t \) is:

\[
Q(t) = 300 \times 5^{-t/120}
\]

Or, equivalently:

\[
Q(t) = 300 \times \left(\frac{1}{5}\right)^{t/120}
\]

Would you like further details, or do you have any questions?

### Related Questions:

1. How does changing the decay time constant \( T \) affect the rate at which the drug decays?
2. What would the function look like if the drug decayed to one-tenth of its value every 120 minutes?
3. How can we convert this exponential decay model into a continuous decay model using \( e \)?
4. What is the half-life of the drug according to this model?
5. How much of the drug remains in the bloodstream after 240 minutes?

**Tip:** Understanding exponential decay is crucial in pharmacokinetics, where it models how drugs dissipate in the body.

Explore how to model drug decay in the bloodstream using exponential decay. Understand the initial quantity, decay rate, and time constant with a detailed step-by-step solution. Suitable for advanced high school students.

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