We are dealing with exponential growth, which is described by the formula:

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

Where:

• $P(t)$ is the population at time $t$,
• $P_0$ is the initial population,
• $k$ is the growth rate constant,
• $t$ is the time in hours,
• $e$ is the base of the natural logarithm (approximately 2.718).

Step 1: Use the initial condition to determine $P_0$

From the problem, we know that the initial population is 100 bacteria, so:

$P_0 = 100$

Thus, the population formula becomes:

$P(t) = 100 \cdot e^{kt}$

Step 2: Use the condition after 1 hour to find $k$

After 1 hour, the population is 210. We can use this information to find the value of $k$:

$P(1) = 210$

Substitute into the formula:

$210 = 100 \cdot e^{k \cdot 1}$

Solve for $e^k$:

$\frac{210}{100} = e^{k}$

$2.1 = e^{k}$

Take the natural logarithm (ln) of both sides:

$\ln(2.1) = k$

$k \approx 0.7419$

Step 3: Write the final formula

Now that we have $k$, the formula for the population after $t$ hours is:

$P(t) = 100 \cdot e^{0.7419t}$

This is the formula for the size of the population at any time $t$.

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Would you like further details or have any questions about the solution?

Here are some related questions you might find interesting:

1. How do we interpret the growth constant $k$?
2. What would the population be after 2 hours?
3. How can we graph the exponential growth of this population?
4. How can we solve for the time it takes for the population to reach a specific number (e.g., 500)?
5. How would the formula change if the growth were instead modeled as logistic growth?

Tip: The exponential growth rate constant $k$ tells you how fast the population is growing. A higher $k$ means faster growth!