\(\mathrm{B(t) = 450(2)^{(t/6)}}\) represents the number of bacteria, in thousands, in a laboratory culture t hours after the start of...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{B(t) = 450(2)^{t/6}}\) represents bacteria population in thousands after t hours
  • Population doubles every n hours (what we need to find)

• What "doubles every n hours" means mathematically:
  • After n additional hours, population is twice as large
  • \(\mathrm{B(t + n) = 2 \times B(t)}\) for any starting time t

2. INFER the solution approach

• Set up the doubling condition as an equation
• Use the given exponential function to create expressions for both sides
• Solve for n by using properties of exponential equations

3. Set up the doubling equation

• Left side: \(\mathrm{B(t + n) = 450(2)^{(t + n)/6}}\)
• Right side: \(\mathrm{2 \times B(t) = 2 \times 450(2)^{t/6}}\)
• Equation: \(\mathrm{450(2)^{(t + n)/6} = 2 \times 450(2)^{t/6}}\)

4. SIMPLIFY by removing common factors

• Divide both sides by 450:
  \(\mathrm{(2)^{(t + n)/6} = 2 \times (2)^{t/6}}\)

• Rewrite 2 as 2^1:
  \(\mathrm{(2)^{(t + n)/6} = (2)^1 \times (2)^{t/6}}\)

• Use exponent rule \(\mathrm{a^m \times a^n = a^{m+n}}\):
  \(\mathrm{(2)^{(t + n)/6} = (2)^{1 + t/6}}\)

5. INFER that equal bases mean equal exponents

• Since both sides have base 2, the exponents must be equal:
  \(\mathrm{(t + n)/6 = 1 + t/6}\)

6. SIMPLIFY to solve for n

• Expand left side: \(\mathrm{t/6 + n/6 = 1 + t/6}\)
• Subtract t/6 from both sides: \(\mathrm{n/6 = 1}\)
• Multiply both sides by 6: \(\mathrm{n = 6}\)

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students often misinterpret what "doubles every n hours" means mathematically. They might think it means \(\mathrm{B(n) = 2 \times B(0)}\) (population at time n equals twice the initial population) instead of the correct interpretation \(\mathrm{B(t + n) = 2 \times B(t)}\) for any time t.

Using \(\mathrm{B(n) = 2 \times B(0)}\), they would set up: \(\mathrm{450(2)^{n/6} = 2 \times 450(2)^0 = 900}\), leading to \(\mathrm{(2)^{n/6} = 2}\), so \(\mathrm{n/6 = 1}\), giving \(\mathrm{n = 6}\). While this accidentally gives the correct answer, it's based on flawed reasoning and wouldn't work for more complex problems.

Second Most Common Error:

Poor INFER reasoning about exponent properties: Some students struggle with the step where \(\mathrm{(2)^{(t + n)/6} = (2)^1 \times (2)^{t/6}}\). They might not recognize that this becomes \(\mathrm{(2)^{1 + t/6}}\), or they might make errors in combining the exponents.

This algebraic confusion can lead them to get stuck and abandon systematic solution, causing them to guess among the answer choices.

The Bottom Line:

This problem tests whether students can correctly translate the concept of "doubling period" into a mathematical equation and then use exponential function properties to solve it. The key insight is that doubling means the ratio between values separated by n hours is always 2:1, regardless of the starting time.