A medication's concentration in a patient's bloodstream decreases exponentially over time.Immediately after the injection, the concentration is 80 mil...

1. TRANSLATE the problem information

• Given information:
  • Model: \(\mathrm{M = C(2)^{rt}}\)
  • At injection (\(\mathrm{t = 0}\)): \(\mathrm{M = 80}\) mg/L
  • Six hours later (\(\mathrm{t = 6}\)): \(\mathrm{M = 20}\) mg/L
  • Need to find: value of \(\mathrm{r}\)

2. INFER the approach

• Since we have two data points and two unknowns (\(\mathrm{C}\) and \(\mathrm{r}\)), we can solve systematically
• Use the initial condition first to find \(\mathrm{C}\), then use the second condition to find \(\mathrm{r}\)

3. SIMPLIFY to find constant C

• At \(\mathrm{t = 0}\): \(\mathrm{M = C(2)^{r \cdot 0}}\)
• \(\mathrm{M = C(2)^{0}}\)
• \(\mathrm{M = C \cdot 1}\)
• \(\mathrm{M = C}\)
• Therefore: \(\mathrm{C = 80}\)

4. TRANSLATE the second condition into an equation

• At \(\mathrm{t = 6}\): \(\mathrm{M = 20}\)
• Substituting into our model: \(\mathrm{20 = 80(2)^{6r}}\)

5. SIMPLIFY the exponential equation

• Divide both sides by 80: \(\mathrm{\frac{20}{80} = (2)^{6r}}\)
• Simplify the fraction: \(\mathrm{\frac{1}{4} = (2)^{6r}}\)

6. INFER how to solve the exponential equation

• We need to express \(\mathrm{\frac{1}{4}}\) as a power of 2
• Since \(\mathrm{2^2 = 4}\), we know that \(\mathrm{\frac{1}{4} = 2^{-2}}\)

7. SIMPLIFY to solve for r

• Now we have: \(\mathrm{(2)^{6r} = 2^{-2}}\)
• Since the bases are equal: \(\mathrm{6r = -2}\)
• Solve for \(\mathrm{r}\): \(\mathrm{r = \frac{-2}{6} = \frac{-1}{3}}\)

Answer: C \(\mathrm{(\frac{-1}{3})}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may misinterpret the exponential model or incorrectly set up the initial conditions. Some students might think \(\mathrm{C}\) represents the final concentration rather than the initial coefficient, or confuse which time corresponds to which concentration value.

This leads to incorrect equations and may cause them to select Choice E \(\mathrm{(1)}\) if they assume \(\mathrm{r}\) must be positive for decay.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students successfully set up the equation \(\mathrm{\frac{1}{4} = 2^{6r}}\) but struggle to convert \(\mathrm{\frac{1}{4}}\) into a power of 2. They might attempt to take logarithms unnecessarily or make arithmetic errors when manipulating the exponential relationship.

This may lead them to select Choice A \(\mathrm{(-1)}\) if they incorrectly conclude that \(\mathrm{\frac{1}{4} = 2^{-6}}\) instead of \(\mathrm{2^{-2}}\).

The Bottom Line:

This problem requires students to bridge real-world exponential decay with abstract mathematical manipulation. The key insight is recognizing that exponential decay problems often require expressing fractions as powers of the base to solve efficiently.