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To determine the equation of an exponential function given a set of data points, we follow a systematic approach. The data is provided as follows:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 7 & 1250 \\ \hline 14 & 500 \\ \hline 21 & 200 \\ \hline 28 & 80 \\ \hline 35 & 32 \\ \hline \end{array} \][/tex]
The goal is to find an exponential function of the form [tex]\( y = ab^x \)[/tex].
### Step 1: Linearizing the Exponential Relation
Firstly, we take the natural logarithm of the y-values to transform the exponential relation into a linear form:
[tex]\[ \begin{array}{|c|c|c|} \hline x & y & \log(y) \\ \hline 7 & 1250 & 7.1309 \\ \hline 14 & 500 & 6.2146 \\ \hline 21 & 200 & 5.2983 \\ \hline 28 & 80 & 4.3820 \\ \hline 35 & 32 & 3.4657 \\ \hline \end{array} \][/tex]
### Step 2: Setting Up the Linear Regression
We assume a linear relationship in the form:
[tex]\[ \log(y) = \log(a) + x \cdot \log(b) \][/tex]
Here, the x-values remain the same, and the y-values are now the natural logarithms of the original y-values.
### Step 3: Performing Linear Regression
Using the linearized form [tex]\(\log(y) = m \cdot x + c\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(c\)[/tex] is the y-intercept:
From our calculations, we find:
[tex]\[ m = -0.1309 \][/tex]
[tex]\[ c = 8.0472 \][/tex]
### Step 4: Back-Transforming to Exponential Form
Solving for [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ \log(b) = m \][/tex]
[tex]\[ \log(a) = c \][/tex]
Thus:
[tex]\[ b = e^{\log(b)} = e^{-0.1309} \approx 0.8773 \][/tex]
[tex]\[ a = e^{\log(a)} = e^{8.0472} \approx 3125 \][/tex]
### Step 5: Writing the Final Equation
With our derived constants [tex]\(a\)[/tex] and [tex]\(b\)[/tex], we can write the exponential function as:
[tex]\[ y = 3125 \cdot (0.8773)^x \][/tex]
This is the exponential function that fits the given data points accurately.
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 7 & 1250 \\ \hline 14 & 500 \\ \hline 21 & 200 \\ \hline 28 & 80 \\ \hline 35 & 32 \\ \hline \end{array} \][/tex]
The goal is to find an exponential function of the form [tex]\( y = ab^x \)[/tex].
### Step 1: Linearizing the Exponential Relation
Firstly, we take the natural logarithm of the y-values to transform the exponential relation into a linear form:
[tex]\[ \begin{array}{|c|c|c|} \hline x & y & \log(y) \\ \hline 7 & 1250 & 7.1309 \\ \hline 14 & 500 & 6.2146 \\ \hline 21 & 200 & 5.2983 \\ \hline 28 & 80 & 4.3820 \\ \hline 35 & 32 & 3.4657 \\ \hline \end{array} \][/tex]
### Step 2: Setting Up the Linear Regression
We assume a linear relationship in the form:
[tex]\[ \log(y) = \log(a) + x \cdot \log(b) \][/tex]
Here, the x-values remain the same, and the y-values are now the natural logarithms of the original y-values.
### Step 3: Performing Linear Regression
Using the linearized form [tex]\(\log(y) = m \cdot x + c\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(c\)[/tex] is the y-intercept:
From our calculations, we find:
[tex]\[ m = -0.1309 \][/tex]
[tex]\[ c = 8.0472 \][/tex]
### Step 4: Back-Transforming to Exponential Form
Solving for [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ \log(b) = m \][/tex]
[tex]\[ \log(a) = c \][/tex]
Thus:
[tex]\[ b = e^{\log(b)} = e^{-0.1309} \approx 0.8773 \][/tex]
[tex]\[ a = e^{\log(a)} = e^{8.0472} \approx 3125 \][/tex]
### Step 5: Writing the Final Equation
With our derived constants [tex]\(a\)[/tex] and [tex]\(b\)[/tex], we can write the exponential function as:
[tex]\[ y = 3125 \cdot (0.8773)^x \][/tex]
This is the exponential function that fits the given data points accurately.
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