A company models the value of a machine t years after purchase by \(\mathrm{V(t) = V_0(0.72)^{t/3}}\). The model can also...

1. TRANSLATE the problem information

• Given: Two equivalent forms of an exponential decay model
  • \(\mathrm{V(t) = V_0(0.72)^{(t/3)}}\)
  • \(\mathrm{V(t) = V_0(1 - p/100)^t}\)
• Find: The value of constant p

2. INFER the approach

• Since both expressions equal \(\mathrm{V(t)}\), they must equal each other
• Key insight: We need to eliminate the variable t to solve for the constant p
• Strategy: Set the expressions equal and use exponent properties

3. SIMPLIFY the equation setup

• Set equal and divide by \(\mathrm{V_0}\):

\(\mathrm{(0.72)^{(t/3)} = (1 - p/100)^t}\)

4. INFER the key algebraic move

• To eliminate t, take both sides to the power of \(\mathrm{1/t}\)
• This uses the rule: \(\mathrm{(a^m)^n = a^{mn}}\)
• Left side: \(\mathrm{[(0.72)^{(t/3)}]^{(1/t)} = (0.72)^{(t/3 \times 1/t)} = (0.72)^{(1/3)}}\)
• Right side: \(\mathrm{[(1 - p/100)^t]^{(1/t)} = (1 - p/100)^1 = 1 - p/100}\)

5. SIMPLIFY to solve for p

• Now we have: \(\mathrm{(0.72)^{(1/3)} = 1 - p/100}\)
• Rearrange: \(\mathrm{p/100 = 1 - (0.72)^{(1/3)}}\)
• Therefore: \(\mathrm{p = 100[1 - (0.72)^{(1/3)}]}\)

6. SIMPLIFY the numerical calculation

• Need to estimate \(\mathrm{(0.72)^{(1/3)}}\)
• Since \(\mathrm{0.9^3 = 0.729}\), which is very close to 0.72
• So \(\mathrm{(0.72)^{(1/3)} \approx 0.896}\) (use calculator)
• \(\mathrm{p = 100(1 - 0.896) = 100(0.104) = 10.4}\)

Answer: B (10)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the strategy of taking both sides to the power \(\mathrm{1/t}\) to eliminate the variable. Instead, they attempt to expand or manipulate the exponents algebraically, which leads to increasingly complex expressions that cannot be solved. This leads to confusion and guessing.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students understand the correct approach but make errors in estimating \(\mathrm{(0.72)^{(1/3)}}\), either by poor approximation techniques or calculator mistakes. They might estimate this value as 0.85 or 0.95, leading to p values around 15 or 5 respectively. This may lead them to select Choice D (15) or guess between the remaining options.

The Bottom Line:

This problem tests whether students can recognize that equivalent exponential expressions can be solved by strategically eliminating variables through fractional exponents—a technique that transforms an impossible algebraic problem into a straightforward numerical calculation.