\(\mathrm{P = 2,400(0.85)^d}\) The equation above models the population, P, of bacteria in a petri dish d days after observation...

1. TRANSLATE the problem information

• Given information:
  • Original equation: \(\mathrm{P = 2,400(0.85)^d}\) (where d is in days)
  • Need to find: equation where time is in hours (h)
• What this tells us: We need to convert the time unit from days to hours in the exponent.

2. INFER the conversion relationship

• Since we're changing from days to hours, we need the basic conversion:
  • 1 day = 24 hours
  • Therefore: d days = d × 24 hours
  • Solving for d: \(\mathrm{d = h/24}\)
• Key insight: The time variable in the exponent needs to change, but the base (0.85) and initial population (2,400) stay the same.

3. TRANSLATE the conversion into the equation

• Substitute \(\mathrm{d = h/24}\) into the original equation:
  • Original: \(\mathrm{P = 2,400(0.85)^d}\)
  • New: \(\mathrm{P = 2,400(0.85)^{(h/24)}}\)
• This matches choice A exactly.

Answer: A

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skills: Students confuse the direction of unit conversion and think that since there are 24 hours in a day, then \(\mathrm{d = 24h}\).

Using \(\mathrm{d = 24h}\) in the original equation gives \(\mathrm{P = 2,400(0.85)^{(24h)}}\), which matches exactly with choice B. This backwards conversion makes the population decay much faster than realistic (24 times faster per hour instead of 1/24 as fast).

This leads them to select Choice B (\(\mathrm{P = 2,400(0.85)^{(24h)}}\)).

Second Most Common Error:

Inadequate INFER reasoning: Students think they need to convert the decay rate (0.85) from daily to hourly instead of converting the time variable.

They might calculate that if 0.85 is the daily factor, the hourly factor should be \(\mathrm{(0.85)^{(1/24)} \approx 0.994}\), leading to \(\mathrm{P = 2,400(0.994)^h}\). This approach misses that exponential models are most easily converted by changing the time variable, not the base.

This may lead them to select Choice C (\(\mathrm{P = 2,400(0.994)^h}\)).

The Bottom Line:

The key insight is recognizing that exponential function time conversion works by changing the time variable in the exponent using the appropriate conversion factor, while keeping the base and coefficient unchanged. Students often get confused about which direction the conversion factor goes or think they need to modify the decay rate instead.