A radioactive isotope has an initial mass of 500 grams and undergoes exponential decay. Scientists determine that the isotope loses...

1. TRANSLATE the problem information

• Given information:
  • Initial mass: 500 grams
  • Loses 8.5% of its mass each year
  • We need a function for remaining mass after t years
• What this tells us: If 8.5% is lost each year, then 100% - 8.5% = 91.5% remains each year

2. INFER the mathematical approach

• This is an exponential decay situation
• For exponential decay, we use the form: \(\mathrm{m(t) = m_0 \times (decay\,factor)^t}\)
• The initial mass \(\mathrm{m_0 = 500}\) grams
• The decay factor = 0.915 (since 91.5% = 0.915 in decimal form)

3. SIMPLIFY to get the final function

• Substituting our values: \(\mathrm{m(t) = 500 \times (0.915)^t}\)

4. APPLY CONSTRAINTS to verify answer choices

• For decay, the base must be between 0 and 1
• Our base is 0.915, which is less than 1 ✓
• This matches choice (B): \(\mathrm{m(t) = 500(0.915)^t}\)

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students might misinterpret "loses 8.5%" and think the decay factor should be 0.085 instead of 0.915.

They forget that the decay factor represents what remains, not what's lost. This conceptual confusion about percentage retention vs. percentage loss leads them to look for functions with 0.085 as the base, but since no answer choice has this, they get confused and may guess randomly.

Second Most Common Error:

Poor INFER reasoning: Students recognize it's exponential but confuse growth with decay, selecting choice (C) \(\mathrm{m(t) = 500(1.085)^t}\) because they see a number close to their percentage calculations.

They think "8.5% means we add 8.5%, so 100% + 8.5% = 108.5% = 1.085" which is backwards reasoning. This may lead them to select Choice C (1.085^t formula).

The Bottom Line:

The key insight is distinguishing between what's lost versus what remains in exponential decay, then correctly applying the standard exponential function form.