The table shows the exponential relationship between the number of days, t, since a radioactive sample was first measured, and...

1. INFER the function structure and initial value

• Given information:
  • Table shows exponential relationship \(\mathrm{m(t) = a(b)^t}\)
  • At \(\mathrm{t = 0}\), \(\mathrm{m(0) = 3.00}\)
  • At \(\mathrm{t = 2}\), \(\mathrm{m(2) = 1.92}\)
  • At \(\mathrm{t = 4}\), \(\mathrm{m(4) = 1.23}\)
• Key insight: In the form \(\mathrm{m(t) = a(b)^t}\), when \(\mathrm{t = 0}\), we get \(\mathrm{m(0) = a(1) = a}\)
• Therefore: \(\mathrm{a = 3.00}\) (the initial value)

2. INFER which answer choices remain viable

• With \(\mathrm{a = 3.00}\), we can eliminate choice (A) immediately since it has \(\mathrm{a = 1.92}\)
• Remaining choices: B, C, and D all have \(\mathrm{a = 3.00}\)

3. SIMPLIFY to find the base b

• Use the second data point: \(\mathrm{m(2) = 1.92}\)
• Substitute into \(\mathrm{m(t) = 3.00(b)^t}\):
  \(\mathrm{3.00(b)^2 = 1.92}\)
• Solve for \(\mathrm{b^2}\):
  \(\mathrm{b^2 = 1.92/3.00 = 0.64}\)
• Take the square root:
  \(\mathrm{b = \sqrt{0.64} = 0.8}\)

4. INFER the verification step

• Check our answer with the third data point:
  \(\mathrm{m(4) = 3.00(0.8)^4 = 3.00(0.4096) = 1.23}\) ✓
• This confirms our function: \(\mathrm{m(t) = 3.00(0.8)^t}\)

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that the coefficient a equals the initial value when \(\mathrm{t = 0}\). They might try to use a different data point first or not understand the connection between the function form and the table structure. This leads to setting up incorrect equations or not eliminating obviously wrong choices immediately. This may lead them to select Choice A \(\mathrm{(1.92(0.8)^t)}\) or causes confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify \(\mathrm{a = 3.00}\) but make arithmetic errors when solving \(\mathrm{3.00(b)^2 = 1.92}\). Common mistakes include incorrectly calculating \(\mathrm{1.92/3.00}\) or taking the square root of \(\mathrm{0.64}\). This may lead them to select Choice B \(\mathrm{(3.00(0.64)^t)}\) by confusing \(\mathrm{b^2}\) with b.

The Bottom Line:

This problem tests whether students truly understand the structure of exponential functions and can connect abstract function notation to concrete data. The key breakthrough is recognizing that the initial condition immediately gives you one parameter, dramatically simplifying the problem.