To solve this problem, we'll break it down into two parts: finding the logarithmic model and using it to make predictions.

A. Finding the best-fitting logarithmic function using least squares

Given the dataTo solve this problem, we'll break it down into two parts: finding the logarithmic model and using it to make predictions.

A. Finding the best-fitting logarithmic function using least squares

Given the

$\text{Percent of 𝛼-terpineol (x)}: 1, 10, 25$ $\text{Days to Adult Stage (y)}: 129, 113, 115$

We aim to find a logarithmic model of the form:

$y(x) = a + b \ln(x)$

We will use the least squares method to find the coefficients $a$ and $b$. The steps include:

1. Transform the $x$ values by taking the natural logarithm.
2. Use the transformed $x$ values and the $y$ values to compute the least-squares regression.
3. Calculate the correlation coefficient ($r$) and square it to get $r^2$.

Let’s go through the calculations.

Step 1: Transform the $x$ values

$\ln(1) = 0, \quad \ln(10) \approx 2.3026, \quad \ln(25) \approx 3.2189$ So, we now have the pairs: $(0, 129), (2.3026, 113), (3.2189, 115)$

Step 2: Use the least squares method

The formulas for the least squares regression coefficients $a$ and $b$ are:

$b = \frac{n\sum(\ln(x) \cdot y) - \sum \ln(x) \cdot \sum y}{n\sum (\ln(x))^2 - (\sum \ln(x))^2}$ $a = \frac{\sum y - b \sum \ln(x)}{n}$

where $n = 3$ (the number of data points). I'll compute these values next.

The best-fitting logarithmic function using least squares is:

$y(x) = 127.9303 - 4.8521 \ln(x)$

The square of the correlation coefficient ($r^2$) is:

$r^2 = 0.8520$

B. Estimate the days to adult stage with a diet of 20% 𝛼-terpineol

We will use the logarithmic model to estimate the number of days to adult stage when the percent of 𝛼-terpineol is 20%.

Substitute $x = 20$ into the equation $y(x) = 127.9303 - 4.8521 \ln(x)$.

Let me compute that for you.When the diet contains 20% 𝛼-terpineol, the estimated number of days to reach the adult stage is approximately 113 days.

Summary:

• Logarithmic function: $y(x) = 127.9303 - 4.8521 \ln(x)$
• $r^2$: 0.8520
• Estimated days for 20% 𝛼-terpineol: 113 days

Do you have any questions or need further details?

Here are some related questions to explore:

1. What is the significance of the correlation coefficient $r^2$?
2. How does the natural logarithmic model compare to a linear model for this data?
3. What other types of models could be considered for this type of biological data?
4. How would the predictions change if the diet percentage was different (e.g., 5% or 50%)?
5. How could you experimentally verify the accuracy of this logarithmic model?

Tip: Always check the correlation coefficient ($r^2$) to assess how well a model fits the data. The closer it is to 1, the better the fit.