A company purchases a new machine for $120,000 in 2022. For accounting purposes, the value of the machine is estimated...

1. TRANSLATE the problem information

• Given information:
  • Initial machine value: \(\mathrm{\$120,000}\) in 2022
  • Machine depreciates by \(\mathrm{15\%}\) of its value each year
  • Need function \(\mathrm{V(t)}\) where \(\mathrm{t}\) = years after purchase

• What this tells us: If the machine loses \(\mathrm{15\%}\) each year, it keeps \(\mathrm{85\%}\) of its value each year.

2. INFER the mathematical model type

• This is exponential decay because the value decreases by a percentage each year
• The general form is \(\mathrm{V(t) = initial\_value \times (retention\_rate)^t}\)
• We need the retention rate, not the depreciation rate

3. TRANSLATE percentages to decimals

• Depreciation rate: \(\mathrm{15\% = 0.15}\)
• Retention rate: \(\mathrm{100\% - 15\% = 85\% = 0.85}\)

4. INFER the correct function structure

• Initial value (a): \(\mathrm{\$120,000}\)
• Base (retention rate): \(\mathrm{0.85}\)
• Time variable: \(\mathrm{t}\) years after purchase
• Function: \(\mathrm{V(t) = 120,000(0.85)^t}\)

5. APPLY CONSTRAINTS by checking answer choices

• Choice (A): \(\mathrm{V(t) = 120,000(0.15)^t}\) uses depreciation rate - incorrect
• Choice (B): \(\mathrm{V(t) = 120,000(0.85)^t}\) uses retention rate - correct
• Choice (C): \(\mathrm{V(t) = 120,000(1.15)^t}\) shows growth - incorrect
• Choice (D): \(\mathrm{V(t) = 120,000(1 - 0.15t)}\) is linear - incorrect

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students directly use the depreciation percentage (\(\mathrm{15\%}\)) as the base of the exponential function without recognizing that exponential decay requires the retention rate.

They read "depreciates by \(\mathrm{15\%}\)" and immediately think the base should be \(\mathrm{0.15}\), missing that if something loses \(\mathrm{15\%}\) each year, it keeps \(\mathrm{85\%}\) each year. The base in exponential decay represents what remains, not what's lost.

This leads them to select Choice A (\(\mathrm{V(t) = 120,000(0.15)^t}\))

Second Most Common Error:

Conceptual confusion about exponential vs linear models: Students might think that "\(\mathrm{15\%}\) per year" means you subtract \(\mathrm{15\%}\) of the original value each year, leading to linear decay rather than exponential.

They might reason: "Each year we lose \(\mathrm{15\%}\) of \(\mathrm{\$120,000}\), so we lose \(\mathrm{\$18,000}\) per year." This linear thinking could lead them toward Choice D (\(\mathrm{V(t) = 120,000(1 - 0.15t)}\)).

The Bottom Line:

The key challenge is understanding that exponential decay uses the retention rate (what's left) as the base, not the decay rate (what's lost). Students must TRANSLATE "loses \(\mathrm{15\%}\) each year" into "keeps \(\mathrm{85\%}\) each year" and then INFER that \(\mathrm{0.85}\) becomes the base of the exponential function.