A radioactive isotope has a mass given by the function \(\mathrm{m(t) = 2,400(2)^{-t/560}}\), where \(\mathrm{m(t)}\) represents the mass in grams...

1. TRANSLATE the problem information

• Given information:
  • Mass function: \(\mathrm{m(t) = 2{,}400(2)^{-t/560}}\)
  • Need to find when mass = one-fourth of original

• What this tells us: We need to find the value of t when the mass equals 1/4 of what it started with

2. TRANSLATE to find the target mass

• Original mass at t = 0: \(\mathrm{m(0) = 2{,}400(2)^{-0/560} = 2{,}400(1) = 2{,}400}\) grams
• One-fourth of original: \(\mathrm{2{,}400 ÷ 4 = 600}\) grams
• We need to solve: \(\mathrm{600 = 2{,}400(2)^{-t/560}}\)

3. SIMPLIFY the equation

• Divide both sides by 2,400: \(\mathrm{600/2{,}400 = (2)^{-t/560}}\)
• This gives us: \(\mathrm{1/4 = (2)^{-t/560}}\)

4. INFER the key insight for solving exponential equations

• To solve this equation, we need both sides to have the same base
• Since \(\mathrm{1/4 = 1/(2^2) = 2^{-2}}\), we can rewrite our equation as: \(\mathrm{2^{-2} = 2^{-t/560}}\)

5. APPLY CONSTRAINTS using exponential equation properties

• When bases are equal: \(\mathrm{2^{-2} = 2^{-t/560}}\), the exponents must be equal
• So: \(\mathrm{-2 = -t/560}\)

6. SIMPLIFY to solve for t

• Multiply both sides by -1: \(\mathrm{2 = t/560}\)
• Multiply both sides by 560: \(\mathrm{t = 2 × 560 = 1{,}120}\) hours

Answer: C (1,120)

Why Students Usually Falter on This Problem

Most Common Error Path:

Missing conceptual knowledge: Students don't know how to convert 1/4 to a power of 2

Many students get stuck at \(\mathrm{1/4 = (2)^{-t/560}}\) because they don't recognize that \(\mathrm{1/4 = 2^{-2}}\). They might try to take logarithms or use other complex methods, leading to confusion and abandoning a systematic approach. This leads to confusion and guessing.

Second Most Common Error:

Weak SIMPLIFY execution: Students make sign errors when solving \(\mathrm{-2 = -t/560}\)

Some students correctly set up \(\mathrm{-2 = -t/560}\) but then solve incorrectly, perhaps forgetting to multiply by the negative sign or making arithmetic mistakes. They might get \(\mathrm{t = -1{,}120}\) or \(\mathrm{t = 280}\). This may lead them to select Choice A (280).

The Bottom Line:

This problem requires recognizing that fractions can be rewritten as negative powers of their denominators' base. Without this insight, students cannot bridge from the rational number 1/4 to the exponential form needed for solving.