Use regression to find an exponential equation that best fits the data above. The equation has form
y

To find an exponential regression equation of the form $y = a \cdot b^x$ that best fits the data given, we can follow these steps:

1. Set Up the Equation for Exponential Regression

An exponential model can be linearized by taking the natural logarithm of both sides: $\ln(y) = \ln(a) + x \ln(b)$ This allows us to treat it as a linear equation in terms of $\ln(y)$, where:

• $\ln(y)$ is the dependent variable,
• $x$ is the independent variable,
• $\ln(a)$ is the intercept,
• $\ln(b)$ is the slope.

2. Transform the Data

For each $y$ value, compute $\ln(y)$ to linearize the data. Here is the data table with $x$ and $\ln(y)$:

$x$	$y$	$\ln(y)$
1	693	$\ln(693) \approx 6.5407$
2	910	$\ln(910) \approx 6.8134$
3	1159	$\ln(1159) \approx 7.0557$
4	1651	$\ln(1651) \approx 7.4085$
5	2190	$\ln(2190) \approx 7.6924$
6	2935	$\ln(2935) \approx 7.9833$

3. Perform Linear Regression on $x$ and $\ln(y)$

Using linear regression techniques, we calculate the best-fit line $\ln(y) = m \cdot x + c$, where $m$ is the slope and $c$ is the intercept.

After calculating or using a regression tool:

• The slope $m \approx 0.2893$,
• The intercept $c \approx 6.248$.

4. Convert Back to the Exponential Model

Since $\ln(y) = 6.248 + 0.2893 \cdot x$, we exponentiate both sides to find $y$ in terms of $x$: $y = e^{6.248} \cdot e^{0.2893x}$ This gives:

• $a \approx e^{6.248} \approx 518.12$,
• $b \approx e^{0.2893} \approx 1.3356$.

Final Answer

The exponential equation that best fits the data is: $y = 518.12 \cdot 1.3356^x$

Would you like further details on the calculation steps or an explanation of any part of this process?

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Related Questions:

1. How can we use this model to predict values beyond $x = 6$?
2. What is the significance of the base $b$ in an exponential equation?
3. How would the model change if the growth rate was decreasing instead?
4. What if we needed a different regression model, such as a polynomial regression?
5. How accurate is the exponential model compared to the actual data?

Tip:

When using exponential regression, always check if the relationship between variables is truly exponential, as other models might sometimes fit better depending on the data pattern.