We are given that the bacteria population grows exponentially, which means the population $y$ after $x$ hours can be modeled by the equation:

$y = y_0 \cdot e^{kx}$ where:

• $y_0$ is the initial population,
• $k$ is the growth rate,
• $x$ is time in hours, and
• $y$ is the population after $x$ hours.

1) Finding the Growth Rate

Given:

• Initial population, $y_0 = 80$ bacteria,
• Population after 8 hours, $y(8) = 160$ bacteria.

Substitute into the exponential growth formula:

$160 = 80 \cdot e^{8k}$

Solve for $k$:

$\frac{160}{80} = e^{8k} \implies 2 = e^{8k}$

Take the natural logarithm of both sides:

$\ln(2) = 8k$

Thus:

$k = \frac{\ln(2)}{8}$

Let’s calculate $k$.

$k \approx \frac{0.693}{8} = 0.0866 \, \text{(rounded to 4 decimal places)}.$

2) Finding the Exponential Equation

Now that we have the growth rate $k \approx 0.0866$, the exponential equation for the bacteria population after $x$ hours is:

$y = 80 \cdot e^{0.0866x}$

This equation gives the population of bacteria at any time $x$, with three decimal places for the growth rate.

Would you like more details on how to interpret the equation, or any clarifications?

Here are five related questions:

1. How would the population change after 24 hours using this equation?
2. What is the doubling time of the bacteria population?
3. How can the equation be adjusted if a different growth rate is observed?
4. What does the growth rate $k$ represent biologically?
5. How would the equation change if the initial population was different?

Tip: When dealing with exponential growth, the natural logarithm is often key to isolating variables like growth rate.