The problem involves exponential growth, where the population of bacteria triples every 12 minutes. To solve this, we'll use the formula for exponential growth:

$P(t) = P_0 \times e^{kt}$

Where:

• $P(t)$ is the population at time $t$,
• $P_0$ is the initial population (which is 1 bacterium in this case),
• $k$ is the growth rate constant,
• $t$ is the time elapsed, and
• $e$ is the base of the natural logarithm (approximately 2.718).

However, since the bacteria triple every 12 minutes, the growth can also be modeled by the formula:

$P(t) = P_0 \times 3^{\frac{t}{12}}$

Here:

• $t$ is the time in minutes,
• 3 represents the factor by which the population grows every 12 minutes.

Now, to find the population after 12 hours, we first convert 12 hours into minutes:

$12 \text{ hours} = 12 \times 60 = 720 \text{ minutes}$

Substituting $t = 720$ minutes into the equation:

$P(720) = 1 \times 3^{\frac{720}{12}} = 3^{60}$

Now, let's compute $3^{60}$.

$P(720) = 3^{60} \approx 2.65614 \times 10^{28}$

So, after 12 hours, the colony will have approximately $2.65614 \times 10^{28}$ bacteria.

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Would you like a more detailed explanation of the steps involved, or do you have any questions about this process?

Here are 5 related questions for further exploration:

1. How does changing the tripling time (e.g., making it 15 minutes) affect the growth?
2. What would happen if the bacteria doubled instead of tripling every 12 minutes?
3. How can we model population decay using exponential functions?
4. What is the concept of the "half-life" in exponential growth/decay problems?
5. How do real-world applications, such as bacteria growth, use this type of exponential model in biology?

Tip: Exponential growth models can grow extremely fast, so even small changes in the growth rate or time can lead to huge differences in outcomes!