A research team models the mass of a bacterial culture, in grams, by the function \(\mathrm{M(t) = 1{,}500 \cdot b^{(t/2)}}\),...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{M(t) = 1,500 \cdot b^{(t/2)}}\) models bacterial mass in grams
  • t represents years after culture started
  • Mass increases by 8% each year
  • Need to find the value of b
• What this tells us: An 8% annual increase means the growth factor per year is 1.08

2. INFER the relationship between time and exponent

• Key insight: In the expression \(\mathrm{b^{(t/2)}}\), when t increases by 1 year, the exponent increases by only 1/2
• This means b doesn't represent the 1-year growth factor directly
• Instead, \(\mathrm{b^{(1/2)}}\) represents the 1-year growth factor

3. TRANSLATE the annual growth into mathematical form

• Since mass grows by 8% annually: \(\mathrm{M(t+1) = 1.08 \cdot M(t)}\)
• Substituting our function:
  • At time t: \(\mathrm{M(t) = 1,500 \cdot b^{(t/2)}}\)
  • At time t+1: \(\mathrm{M(t+1) = 1,500 \cdot b^{((t+1)/2)}}\)
    \(\mathrm{= 1,500 \cdot b^{(t/2 + 1/2)}}\)
    \(\mathrm{= 1,500 \cdot b^{(t/2)} \cdot b^{(1/2)}}\)

4. INFER the equation to solve

• From the growth relationship: \(\mathrm{1,500 \cdot b^{(t/2)} \cdot b^{(1/2)} = 1.08 \cdot (1,500 \cdot b^{(t/2)})}\)
• Dividing both sides by \(\mathrm{1,500 \cdot b^{(t/2)}}\): \(\mathrm{b^{(1/2)} = 1.08}\)

5. SIMPLIFY to find b

• Square both sides: \(\mathrm{b = (1.08)^2}\)
• Calculate: \(\mathrm{b = 1.1664}\) (use calculator)

Answer: D. 1.1664

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students assume b directly represents the annual growth factor of 1.08, missing that the exponent is t/2, not t.

They think: "8% annual growth means \(\mathrm{b = 1.08}\)" and select Choice C (1.08) without analyzing how the exponent structure affects the interpretation of b.

Second Most Common Error Path:

Poor TRANSLATE reasoning: Students correctly identify that \(\mathrm{b^{(1/2)} = 1.08}\) but make arithmetic errors when computing \(\mathrm{(1.08)^2}\).

Common calculation mistakes include getting 1.16 instead of 1.1664, leading them to select Choice B (1.04) as the closest option, or making other computational errors.

The Bottom Line:

This problem tests whether students can distinguish between the parameter in an exponential function and the actual growth rate, especially when the exponent involves a coefficient other than 1. The key insight is recognizing that b represents growth over the time period that makes the exponent equal to 1.