\(\mathrm{M(d) = 7,200(0.91)^d}\)The function M above models the mass, in grams, of a radioactive sample remaining d days after the...

1. TRANSLATE the problem information

• Given information:
  • Original function: \(\mathrm{M(d)} = 7{,}200(0.91)^\mathrm{d}\) (mass after d days)
  • Need: \(\mathrm{m(h)}\) function (mass after h hours)
• The problem asks us to convert from a days-based function to an hours-based function

2. INFER the conversion strategy

• The key insight: Both functions model the same radioactive decay, just with different time units
• We need to express the time relationship between days and hours
• Since the exponential base (0.91) represents the daily decay rate, we must convert hours to days

3. TRANSLATE the time conversion

• Time relationship: \(24\text{ hours} = 1\text{ day}\)
• Therefore: \(\mathrm{h}\text{ hours} = \mathrm{h}/24\text{ days}\)
• Mathematical relationship: \(\mathrm{d} = \mathrm{h}/24\)

4. SIMPLIFY by substitution

• Substitute \(\mathrm{d} = \mathrm{h}/24\) into the original function:
• \(\mathrm{M(d)} = \mathrm{M(h/24)} = 7{,}200(0.91)^{(\mathrm{h}/24)}\)
• Therefore: \(\mathrm{m(h)} = 7{,}200(0.91)^{(\mathrm{h}/24)}\)

Answer: C. \(\mathrm{m(h)} = 7{,}200(0.91)^{(\mathrm{h}/24)}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students often confuse which direction the time conversion should go. They might think "since we're going from days to hours, and hours are smaller units, we should multiply by 24" rather than recognizing that h hours equals h/24 days.

This incorrect reasoning leads them to write \(\mathrm{d} = 24\mathrm{h}\), giving them \(\mathrm{m(h)} = 7{,}200(0.91)^{(24\mathrm{h})}\), which matches Choice B.

Second Most Common Error:

Poor TRANSLATE reasoning: Students may correctly identify that 24 is involved but apply it to the wrong part of the function. They might think the initial amount or the base needs to be adjusted by the factor of 24.

This could lead them to select Choice A (\(\mathrm{m(h)} = 7{,}200(0.91/24)^\mathrm{h}\)) or Choice D (\(\mathrm{m(h)} = (7{,}200/24)(0.91)^\mathrm{h}\)), both of which incorrectly modify the wrong components of the exponential function.

The Bottom Line:

This problem tests whether students can maintain the logical structure of exponential functions while correctly handling unit conversions. The key insight is that time conversions affect the exponent, not the base or coefficient, and that converting to smaller time units means dividing, not multiplying, in the exponent.