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Let's break down the problem and solve it step by step.

### Given Data:
The biological model of toxin concentration over time is given by the formula:
\[
f(x) = b \cdot a^x
\]
where:
- \( f(x) \) is the concentration in ppm at time \( x \) (measured in days).
- \( b \) and \( a \) are constants we need to determine.

The data table provides time \( x \) (in days after the first measurement) and corresponding concentrations in ppm:

| Time (days) \( x \) | 0 | 4 | 8 | 12 | 16 | 20 | 24 | 28 |
|----------------------|---|---|---|----|----|----|----|----|
| Concentration (ppm)   | 278 | 206 | 138 | 110 | 81 | 58 | 49 | 32 |

### Part (a) - Determine \( a \) and \( b \) using regression:

We need to fit the exponential model \( f(x) = b \cdot a^x \) to the given data. Taking the natural logarithm of both sides will linearize the equation:
\[
\ln(f(x)) = \ln(b) + x \cdot \ln(a)
\]
So we can perform a linear regression on \( \ln(f(x)) \) against \( x \).

Let's proceed to the regression analysis.

### Part (b) - Time when concentration is under 5 ppm:

We are asked to determine when the concentration drops below 5 ppm. From the model \( f(x) = b \cdot a^x \), we set \( f(x) = 5 \) and solve for \( x \):
\[
5 = b \cdot a^x
\]
\[
x = \frac{\ln(5) - \ln(b)}{\ln(a)}
\]

### Part (c) - Half-life of the toxin:

The half-life \( T_{1/2} \) is the time it takes for the concentration to be halved. So, we need to solve for \( T_{1/2} \) from:
\[
\frac{f(0)}{2} = f(T_{1/2}) = b \cdot a^{T_{1/2}}
\]
Let me calculate these values based on the data provided and get back to you with the results.From the regression analysis, the values of \( b \) and \( a \) in the model \( f(x) = b \cdot a^x \) are:

- \( b \approx 276.62 \)
- \( a \approx 0.925 \)

### Next steps:
Now, let's move to part (b) to calculate the time when the concentration drops below 5 ppm, and then determine the half-life for part (c).

I'll calculate these values now.### Results:

- **Part (b):** The concentration will drop below 5 ppm approximately **51.60 days** after the first measurement.
- **Part (c):** The half-life of the toxin in the lake is approximately **8.91 days**.

Would you like more details on any of the calculations? Here are a few related questions:

1. How is the regression model determined for exponential decay?
2. How can the half-life formula be derived from the model?
3. What is the impact of small variations in \(a\) on the toxin decay rate?
4. How accurate are the regression predictions compared to the actual data?
5. What are the assumptions behind using an exponential decay model?

**Tip:** When dealing with exponential decay, small changes in the base \(a\) can significantly impact the time predictions for reaching specific concentrations.

A biologist measures the concentration of a specific toxin in a polluted lake over time. The concentration data is modeled as an exponential decay function f(x) = b * a^x. The user is tasked to (a) find the values of 'a' and 'b' using regression, (b) determine how long it takes for the concentration to drop below 5 ppm, and (c) calculate the half-life of the toxin.

Solve an exponential decay problem involving the concentration of a toxin in a polluted lake. Learn how to use regression to find the parameters 'a' and 'b', calculate when the concentration drops below 5 ppm, and determine the toxin's half-life. This problem includes exponential decay models and logarithmic transformations. Suitable for advanced high school students (Grades 11-12).

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