The function \(\mathrm{f(t) = 60,000(2)^{(t/410)}}\) gives the number of bacteria in a population t minutes after an initial observation. How...

1. TRANSLATE the problem information

• Given function: \(\mathrm{f(t) = 60,000(2)^{t/410}}\)
• Need to find: Time for the population to double
• What "double" means: \(\mathrm{f(t) = 2 \times f(0)}\)

2. INFER the key relationship

• In exponential functions with base 2, the expression \(\mathrm{2^{t/410}}\) doubles when the exponent increases by 1
• Since the base is 2, we need \(\mathrm{2^{t/410} = 2^1}\)
• This gives us \(\mathrm{t/410 = 1}\)

3. SIMPLIFY to find the answer

• From \(\mathrm{t/410 = 1}\)
• Multiply both sides by 410: \(\mathrm{t = 410}\)

Answer: 410 minutes

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the connection between exponential doubling and the exponent increasing by 1. Instead, they might try to set up a complex equation like:

\(\mathrm{60,000(2)^{t/410} = 2 \times 60,000(2)^{t/410}}\)

This equation has no solution since both sides are identical, leading to confusion and guessing.

Second Most Common Error:

Poor TRANSLATE reasoning: Students might misunderstand what "doubling time" means and try to find when \(\mathrm{f(t) = 2t}\) or some other incorrect interpretation, rather than when \(\mathrm{f(t) = 2 \times f(0)}\).

This leads them to set up the wrong equation entirely, causing them to get stuck and randomly select an answer.

The Bottom Line:

This problem tests whether students understand the fundamental property of exponential functions with base 2: the output doubles when the exponent increases by 1. Missing this key insight makes the problem much harder than it needs to be.