\(\mathrm{P(t) = 260(1.04)^{(6/4)t}}\) The function P models the population, in thousands, of a certain city t years after 2003. According...

1. TRANSLATE the problem information

• Given function: \(\mathrm{P(t) = 260(1.04)^{(6/4)t}}\) where t is years after 2003
• Population increases by 4% every n months
• Need to find: value of n

2. SIMPLIFY the exponent

• \(\mathrm{(6/4)t = (3/2)t = 1.5t}\)
• So the function becomes: \(\mathrm{P(t) = 260(1.04)^{1.5t}}\)

3. INFER the growth pattern

• In exponential functions, the base 1.04 means 4% growth occurs when the exponent increases by 1
• The exponent here is \(\mathrm{1.5t}\)
• So we need to find: when does 1.5t increase by 1?

4. SIMPLIFY to find the time interval

• For the exponent to increase by 1: \(\mathrm{1.5(t_2 - t_1) = 1}\)
• Solving: \(\mathrm{t_2 - t_1 = 1/1.5 = 2/3}\) years
• This means 4% growth occurs every 2/3 years

5. TRANSLATE years to months

• \(\mathrm{2/3\text{ years} \times 12\text{ months/year} = 8\text{ months}}\)

Answer: A. 8

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER reasoning: Students may incorrectly think that since the base is 1.04 (representing 4% growth), and there are 12 months in a year, the answer is 12 months. They miss the crucial step of analyzing when the exponent actually increases by 1.

This may lead them to select Choice B (12).

Second Most Common Error:

Incomplete SIMPLIFY execution: Students correctly identify that they need to find when the exponent increases by 1, but make algebraic errors when solving \(\mathrm{1.5(t_2 - t_1) = 1}\), possibly getting \(\mathrm{t_2 - t_1 = 1.5}\) instead of 2/3.

This leads to confusion about the time period and may cause them to guess among the remaining choices.

The Bottom Line:

This problem requires understanding that exponential growth rates are tied to unit changes in the exponent, not simply the presence of a growth factor in the base.