An initial investment of $2,500 grows in value over several years. The value, V, of the investment after t years...

1. TRANSLATE the problem information

• Given information:
  • Investment formula: \(\mathrm{V = P(3)^{(t/k)}}\)
  • Initial investment: $2,500
  • Value after 6 years: $7,500
  • Need to find: k

2. INFER what the initial condition tells us

• At \(\mathrm{t = 0}\) (initial time), the investment value should equal the initial investment
• This means: \(\mathrm{V = P(3)^{(0/k)} = P(1) = P}\)
• Therefore: \(\mathrm{P = \$2,500}\)

3. TRANSLATE the 6-year condition into an equation

• After 6 years: \(\mathrm{V = \$7,500}\)
• Substitute into formula: \(\mathrm{7,500 = 2,500(3)^{(6/k)}}\)

4. SIMPLIFY to isolate the exponential term

• Divide both sides by 2,500: \(\mathrm{3 = 3^{(6/k)}}\)

5. INFER the relationship from exponential equality

• Since we have the same base (3) on both sides with equal values
• The exponents must be equal: \(\mathrm{1 = 6/k}\)

6. SIMPLIFY to solve for k

• From \(\mathrm{1 = 6/k}\), multiply both sides by k: \(\mathrm{k = 6}\)

Answer: D) 6

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students often struggle to connect the "initial investment" with the mathematical meaning of P in the formula. They might try to work with the 6-year data point first without establishing what P equals, leading to an equation with two unknowns that they can't solve systematically. This leads to confusion and guessing.

Second Most Common Error:

Inadequate INFER reasoning: Some students correctly find \(\mathrm{P = 2,500}\) but then get confused about how to handle the equation \(\mathrm{3 = 3^{(6/k)}}\). They might try complex logarithmic approaches or fail to recognize that equal bases with equal values means equal exponents. This may lead them to select Choice A (1/6) by incorrectly inverting the final relationship.

The Bottom Line:

This problem tests whether students can systematically use given conditions to find unknown parameters in exponential models. The key insight is recognizing that the initial condition gives you one constant, which then allows you to use the second condition to find the other constant.