A scientist is studying a radioactive element. The mass M, in grams, of the element remaining after t years is...

1. TRANSLATE the problem information

• Given information:
  • Mass function: \(\mathrm{M(t) = 150(2)^{-t/95}}\)
  • Need to find: half-life (time for mass to become half the initial amount)

• What this tells us: We need the time when the mass equals half of the starting mass

2. TRANSLATE the half-life definition into math

• Find initial mass: \(\mathrm{M(0) = 150(2)^{-0/95} = 150(1) = 150}\) grams
• Half of initial mass: \(\mathrm{150 ÷ 2 = 75}\) grams
• Set up equation: \(\mathrm{M(t) = 75}\)

3. INFER the solution strategy

• We have: \(\mathrm{75 = 150(2)^{-t/95}}\)
• Strategy: Isolate the exponential term, then use the property that equal bases mean equal exponents

4. SIMPLIFY to isolate the exponential

• Divide both sides by 150: \(\mathrm{75/150 = (2)^{-t/95}}\)
• This gives us: \(\mathrm{1/2 = (2)^{-t/95}}\)

5. INFER how to handle the exponential equation

• Express 1/2 as a power of 2: Since \(\mathrm{1/2 = 2^{-1}}\)
• Now we have: \(\mathrm{2^{-1} = (2)^{-t/95}}\)
• With equal bases, the exponents must be equal: \(\mathrm{-1 = -t/95}\)

6. SIMPLIFY to solve for t

• Multiply both sides by -95: \(\mathrm{(-1) × (-95) = t}\)
• Therefore: \(\mathrm{t = 95}\) years

Answer: B) 95

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students might confuse what "half-life" means and incorrectly think they need to find when M(t) equals half of some other value (like half of the exponent 95), rather than half of the initial mass.

This confusion leads them to guess randomly or potentially select Choice A (47.5) if they think half-life means "half of 95."

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly set up M(t) = 75 but make algebraic errors when manipulating the exponential equation, such as incorrectly handling the negative exponent or making arithmetic mistakes in the division.

This leads to confusion and incorrect calculations, potentially causing them to select Choice C (150) if they confuse the initial mass with the answer.

The Bottom Line:

This problem requires clear understanding of what half-life means in context, combined with solid exponential equation solving skills. The key insight is recognizing that half-life problems always start with finding half the initial value, then solving when the function equals that amount.