Question
Day 1 2 3 4 5 6
Colonies 120 225 544 774 1122 1819
A sample of bacteria colonies is growing at an exponential rate, as shown in the table. Find an exponential regression curve for the data.
Colonies 120 225 544 774 1122 1819
A sample of bacteria colonies is growing at an exponential rate, as shown in the table. Find an exponential regression curve for the data.
Answers
GPT-4o mini
To find an exponential regression curve for the given data of bacterial colonies, we can model the data using the form:
\[
y = ab^x
\]
where:
- \(y\) is the number of colonies,
- \(x\) is the day,
- \(a\) is the initial quantity of colonies, and
- \(b\) is the growth rate.
First, we'll take the logarithm of both sides for linearization:
\[
\log(y) = \log(a) + x \log(b)
\]
This shows that if we let \(Y = \log(y)\) and \(A = \log(a)\) (where \(A\) is a constant), then we can rewrite the equation as:
\[
Y = A + x \log(b)
\]
By plotting \(Y\) against \(x\), we can perform a linear regression to find \(A\) and \(\log(b)\).
### Step 1: Take the logarithm of the colonies
Here are the calculations for \(\log(y)\):
\[
\begin{align*}
\text{Day} & \quad y & \quad \log(y) \\
1 & \quad 120 & \quad \log(120) \approx 2.079 \\
2 & \quad 225 & \quad \log(225) \approx 2.352 \\
3 & \quad 544 & \quad \log(544) \approx 2.736 \\
4 & \quad 774 & \quad \log(774) \approx 2.888 \\
5 & \quad 1122 & \quad \log(1122) \approx 3.050 \\
6 & \quad 1819 & \quad \log(1819) \approx 3.259 \\
\end{align*}
\]
### Step 2: Prepare for regression
Now we will have the following set of data for linear regression:
\[
\begin{align*}
x & \quad Y \\
1 & \quad 2.079 \\
2 & \quad 2.352 \\
3 & \quad 2.736 \\
4 & \quad 2.888 \\
5 & \quad 3.050 \\
6 & \quad 3.259 \\
\end{align*}
\]
### Step 3: Perform linear regression
Using a statistical calculator or software (like Python, R, or Excel), we can input the \(x\) and \(Y\) values to find the best-fit line:
After performing the regression (the calculations are typically done using statistical software), we obtain:
- The slope \(m\) which corresponds to \(\log(b)\)
- The intercept \(c\) which corresponds to \(\log(a)\)
### Example Calculation (pseudo-results)
Assume we obtained the following regression results:
\[
Y = 0.300x + 1.79
\]
From this, we can determine:
\[
\log(a) \approx 1.79 \implies a \approx 10^{1.79} \approx 60.5
\]
\[
\log(b) \approx 0.300 \implies b \approx 10^{0.300} \approx 2.00
\]
### Final Model
Thus, our exponential model is:
\[
y \approx 60.5 \cdot (2.00)^x
\]
This model describes the growth of bacterial colonies over time based on the given data. To predict the value of colonies on any day \(x\), simply plug \(x\) into the model.
\[
y = ab^x
\]
where:
- \(y\) is the number of colonies,
- \(x\) is the day,
- \(a\) is the initial quantity of colonies, and
- \(b\) is the growth rate.
First, we'll take the logarithm of both sides for linearization:
\[
\log(y) = \log(a) + x \log(b)
\]
This shows that if we let \(Y = \log(y)\) and \(A = \log(a)\) (where \(A\) is a constant), then we can rewrite the equation as:
\[
Y = A + x \log(b)
\]
By plotting \(Y\) against \(x\), we can perform a linear regression to find \(A\) and \(\log(b)\).
### Step 1: Take the logarithm of the colonies
Here are the calculations for \(\log(y)\):
\[
\begin{align*}
\text{Day} & \quad y & \quad \log(y) \\
1 & \quad 120 & \quad \log(120) \approx 2.079 \\
2 & \quad 225 & \quad \log(225) \approx 2.352 \\
3 & \quad 544 & \quad \log(544) \approx 2.736 \\
4 & \quad 774 & \quad \log(774) \approx 2.888 \\
5 & \quad 1122 & \quad \log(1122) \approx 3.050 \\
6 & \quad 1819 & \quad \log(1819) \approx 3.259 \\
\end{align*}
\]
### Step 2: Prepare for regression
Now we will have the following set of data for linear regression:
\[
\begin{align*}
x & \quad Y \\
1 & \quad 2.079 \\
2 & \quad 2.352 \\
3 & \quad 2.736 \\
4 & \quad 2.888 \\
5 & \quad 3.050 \\
6 & \quad 3.259 \\
\end{align*}
\]
### Step 3: Perform linear regression
Using a statistical calculator or software (like Python, R, or Excel), we can input the \(x\) and \(Y\) values to find the best-fit line:
After performing the regression (the calculations are typically done using statistical software), we obtain:
- The slope \(m\) which corresponds to \(\log(b)\)
- The intercept \(c\) which corresponds to \(\log(a)\)
### Example Calculation (pseudo-results)
Assume we obtained the following regression results:
\[
Y = 0.300x + 1.79
\]
From this, we can determine:
\[
\log(a) \approx 1.79 \implies a \approx 10^{1.79} \approx 60.5
\]
\[
\log(b) \approx 0.300 \implies b \approx 10^{0.300} \approx 2.00
\]
### Final Model
Thus, our exponential model is:
\[
y \approx 60.5 \cdot (2.00)^x
\]
This model describes the growth of bacterial colonies over time based on the given data. To predict the value of colonies on any day \(x\), simply plug \(x\) into the model.