\(\mathrm{f(x) = 5,470(0.64)^{(x/12)}}\)The function f gives the value, in dollars, of a certain piece of equipment after x months of...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{f(x) = 5,470(0.64)^{(x/12)}}\) gives equipment value after x months
  • Value decreases each year by p% of preceding year's value
  • Need to find p

2. INFER the time period connection

• The exponent has \(\mathrm{x/12}\), which means the base 0.64 applies over 12-month periods
• Since \(\mathrm{12\,months = 1\,year}\), the function shows yearly decay
• This matches the problem asking for yearly percentage decrease

3. INFER what happens over one year

• At start (\(\mathrm{x = 0}\)): \(\mathrm{f(0) = 5,470(0.64)^0 = 5,470}\)
• After 1 year (\(\mathrm{x = 12}\)): \(\mathrm{f(12) = 5,470(0.64)^1 = 5,470 \times 0.64}\)
• The equipment retains \(\mathrm{0.64}\) (or 64%) of its value after one year

4. INFER the percentage decrease

• If 64% remains, then \(\mathrm{100\% - 64\% = 36\%}\) was lost
• Therefore, \(\mathrm{p = 36}\)

Answer: C. 36

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students see 0.64 in the function and directly interpret this as the percentage decrease, thinking "the equipment decreases by 64% each year."

They fail to recognize that 0.64 represents the remaining fraction, not the lost fraction. This leads them to select Choice D (64).

Second Most Common Error:

Incomplete INFER reasoning: Students may recognize that they need to find a yearly rate but get confused about the relationship between the exponent structure (\(\mathrm{x/12}\)) and the time period.

Without clearly connecting that k = 12 means the base applies over 12-month periods, they might try random calculations or guess among the choices.

The Bottom Line:

The key insight is distinguishing between the decay factor (what remains) and the decay percentage (what's lost). In exponential decay, r represents what you keep, while the percentage decrease is what you lose: \(\mathrm{100(1-r)\%}\).