The table shows the relationship between the time t, in hours, since a medication was administered, and the concentration \(\mathrm{C(t)}\),...

1. TRANSLATE the problem information

• Given information:
  • Table with time \(\mathrm{t}\) (hours) and concentration \(\mathrm{C(t)}\) (mg/L)
  • \(\mathrm{t = 0}\): \(\mathrm{C(t) = 48.0}\)
  • \(\mathrm{t = 3}\): \(\mathrm{C(t) = 38.4}\)
  • \(\mathrm{t = 6}\): \(\mathrm{C(t) = 30.72}\)
  • The relationship is exponential

2. INFER the structure of the exponential function

• For exponential functions of the form \(\mathrm{C(t) = A \cdot b^{(something\ with\ t)}}\), we can use the initial condition to find A
• At \(\mathrm{t = 0}\), any exponential expression equals 1, so \(\mathrm{C(0) = A \cdot 1 = A}\)
• Therefore: \(\mathrm{A = 48}\)

3. INFER the decay pattern by examining ratios

• Check how the concentration changes over equal time intervals
• From \(\mathrm{t = 0}\) to \(\mathrm{t = 3}\): ratio = \(\mathrm{38.4/48 = 0.8}\) (use calculator)
• From \(\mathrm{t = 3}\) to \(\mathrm{t = 6}\): ratio = \(\mathrm{30.72/38.4 = 0.8}\) (use calculator)
• Key insight: Every 3 hours, the concentration is multiplied by 0.8

4. INFER the correct exponent form

• Since the multiplication factor of 0.8 occurs every 3 hours (not every hour), the exponent should represent "number of 3-hour periods"
• Number of 3-hour periods = \(\mathrm{t/3}\)
• Therefore: \(\mathrm{C(t) = 48(0.8)^{(t/3)}}\)

5. SIMPLIFY to verify the function works

• Check \(\mathrm{t = 0}\):
  \(\mathrm{C(0) = 48(0.8)^{(0/3)}}\)
  \(\mathrm{= 48(0.8)^0}\)
  \(\mathrm{= 48(1)}\)
  \(\mathrm{= 48}\) ✓
• Check \(\mathrm{t = 3}\):
  \(\mathrm{C(3) = 48(0.8)^{(3/3)}}\)
  \(\mathrm{= 48(0.8)^1}\)
  \(\mathrm{= 48(0.8)}\)
  \(\mathrm{= 38.4}\) ✓
• Check \(\mathrm{t = 6}\):
  \(\mathrm{C(6) = 48(0.8)^{(6/3)}}\)
  \(\mathrm{= 48(0.8)^2}\)
  \(\mathrm{= 48(0.64)}\)
  \(\mathrm{= 30.72}\) ✓ (use calculator)

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students recognize they need an exponential function with \(\mathrm{A = 48}\), but fail to realize that the decay factor applies every 3 hours, not every hour. They might think since the base is 0.8, the function should be \(\mathrm{C(t) = 48(0.8)^t}\).

If they check this:
\(\mathrm{C(3) = 48(0.8)^3}\)
\(\mathrm{= 48(0.512)}\)
\(\mathrm{= 24.58}\), which doesn't match the given 38.4.

This confusion about the time interval often leads them to select Choice A (\(\mathrm{C(t) = 48(0.80)^t}\)) or causes them to get stuck and guess.

Second Most Common Error:

Poor TRANSLATE reasoning: Students might focus on individual data points rather than recognizing the pattern in ratios between consecutive values. They may attempt to work backwards from each answer choice without understanding the underlying exponential relationship.

This leads to inefficient checking and potential calculation errors, often resulting in abandoning systematic solution and guessing.

The Bottom Line:

The key challenge is recognizing that exponential functions can have exponents other than just "\(\mathrm{t}\)" - in this case, the time interval for the decay factor determines that we need \(\mathrm{t/3}\) in the exponent.