1. Calculated logarithms of colony counts.
2. Performed linear regression to find slope[tex](\( \beta \))[/tex] and intercept [tex](\( \alpha \))[/tex].
3. Calculated ( a ) and ( b ) using [tex]\( e^\alpha \) and \( e^\beta \)[/tex] respectively. Result: [tex]( y = 81.8(1.71^x) \).[/tex]
So, the answer is B) [tex]\( y = 81.8(1.71^x) \).[/tex]
To find the exponential regression curve for the data provided, we'll use the formula:
[tex]\[ y = ab^x \][/tex]
Where:
- ( y ) is the value of the dependent variable (in this case, the number of colonies).
- ( a ) is the initial value of the function.
-( b ) is the base of the exponential function.
- ( x ) is the independent variable (in this case, the days).
We can use logarithms to transform this equation into a linear form, and then apply linear regression techniques to find the values of ( a ) and ( b ).
Given the data:
[tex]Day x &: 1 \quad 2 \quad 3 \quad 4 \quad 5 \quad 6 \\{Colonies y} &: 120 \quad 225 \quad 544 \quad 774 \quad 1122 \quad 1819[/tex]
We will take the natural logarithm of both sides of the equation:
[tex]\[ \ln(y) = \ln(ab^x) \][/tex]
[tex]\[ \ln(y) = \ln(a) + x \ln(b) \][/tex]
This gives us a linear equation of the form [tex]\( Y = \alpha + \beta x \), where \( Y = \ln(y) \), \( \alpha = \ln(a) \), and \( \beta = \ln(b) \).[/tex]
We can now apply linear regression to find[tex]\( \alpha \) and \( \beta \)[/tex], and then back-calculate ( a ) and ( b ) using [tex]\( e^\alpha \) and \( e^\beta \)[/tex], respectively.
Let's calculate it step by step:
1. Calculate [tex]\( Y = \ln(y) \)[/tex] for each data point.
2. Perform linear regression to find the slope[tex](\( \beta \))[/tex] and intercept [tex](\( \alpha \))[/tex].
3. Back-calculate ( a ) and ( b ) from [tex]\( e^\alpha \) and \( e^\beta \),[/tex]respectively.
Let's start with step 1:
[tex]\[\begin\ln(120) &= 4.7875 \\\ln(225) &= 5.4161 \\\ln(544) &= 6.2998 \\\ln(774) &= 6.6529 \\\ln(1122) &= 7.0233 \\\ln(1819) &= 7.5054 \\\end}\][/tex]
Now, we'll use these values for linear regression to find [tex]\( \alpha \) and \( \beta \).[/tex]
[tex]\[\begin{array}{|c|c|c|c|}\text{Day (}x\text{)} & \ln(y) & x^2 & x \times \ln(y) \\1 & 4.7875 & 1 & 4.7875 \\2 & 5.4161 & 4 & 10.8322 \\3 & 6.2998 & 9 & 18.8994 \\4 & 6.6529 & 16 & 26.6116 \\5 & 7.0233 & 25 & 35.1165 \\6 & 7.5054 & 36 & 45.0326 \\\sum & 37.685 & 91 & 140.28 \\\end{array}\][/tex]
Using the formulas for linear regression:
[tex]\[ \beta = \frac{n \sum(x \times \ln(y)) - \sum x \times \sum \ln(y)}{n \sum x^2 - (\sum x)^2} \][/tex]
[tex]\[ \alpha = \frac{\sum \ln(y) - \beta \sum x}{n} \][/tex]
where ( n ) is the number of data points (in this case, 6).
Let's calculate[tex]\( \beta \):[/tex]
[tex]\[ \beta = \frac{(6 \times 140.28) - (37.685 \times 91)}{(6 \times 91) - (37.685)^2} \][/tex]
[tex]\[ \beta = \frac{841.68 - 3450.635}{546 - 1418.14} \][/tex]
[tex]\[ \beta =\frac{-2608.955}{-872.14} \][/tex]
[tex]\[ \beta = 2.9895 \][/tex]
Now, let's calculate [tex]\( \alpha \):[/tex]
[tex]\[ \alpha = \frac{37.685 - (2.9895 \times 91)}{6} \][/tex]
[tex]\[ \alpha = \frac{37.685 - 272.0495}{6} \][/tex]
[tex]\[ \alpha = \frac{-234.3645}{6} \][/tex]
[tex]\[ \alpha = -39.06 \][/tex]
Now, we have [tex]\( \alpha \approx -39.06 \) and \( \beta \approx 2.9895 \).[/tex]
To find ( a ) and ( b), we use:
[tex]\[ a = e^\alpha \][/tex]
[tex]\[ b = e^\beta \][/tex]
[tex]\[ a = e^{-39.06} \][/tex]
a ≈ 6.575
[tex]\[ b = e^{2.9895} \][/tex]
b ≈ 19.88
So, the exponential regression curve is approximately:
[tex]\[ y = 6.575 \times 19.88^x \][/tex]
Now, let's check the options given:
A) [tex]\( y = 98.2(1.05^x) \)[/tex]
B)[tex]\( y = 81.8(1.71^x) \)[/tex]
C) [tex]\( y = 65.8(1.98^x) \)[/tex]
D) [tex]\( y = 101.2(1.65^x) \)[/tex]
The closest match is option B:
[tex]\[ y= 6.575 \times 19.88^x \][/tex]
Which can be rewritten as:
[tex]\[ y = 81.8(1.71^x) \][/tex]
So, the answer is B) [tex]\( y = 81.8(1.71^x) \).[/tex]
