To find the exponential function $f(x) = a \cdot b^x$ that best fits the data provided, we will follow these steps:

1. Formulate the model: Assume $f(x) = a \cdot b^x$.
2. Apply logarithmic transformation: To linearize the exponential model, we can take the natural logarithm of both sides, giving us: $\ln(y) = \ln(a) + x \cdot \ln(b)$ This transformation allows us to use linear regression on $(x, \ln(y))$ to find $\ln(a)$ and $\ln(b)$.
3. Calculate values of $\ln(y)$ for each $x$ value: $x = 1, y = 1 \Rightarrow \ln(1) = 0$ $x = 2, y = 6 \Rightarrow \ln(6) \approx 1.79$ $x = 3, y = 14 \Rightarrow \ln(14) \approx 2.64$ $x = 4, y = 37 \Rightarrow \ln(37) \approx 3.61$ $x = 5, y = 115 \Rightarrow \ln(115) \approx 4.74$
4. Perform linear regression on $(x, \ln(y))$ to determine the best fit line, from which we can extract $\ln(a)$ and $\ln(b)$.

Let's proceed with this approach to determine the values of $a$ and $b$.The best-fit exponential function for $f(x)$ based on the data provided is:

$f(x) \approx 0.434 \cdot 3.098^x$

This model fits the data quite well, with an $R^2$ value of approximately $0.982$, indicating a strong correlation between the model and the data.

Would you like further details on the calculations or have any questions?

Related Questions

1. How can I use this exponential function to estimate values of $y$ for other $x$-values?
2. What does the $R^2$ value indicate in terms of the model's accuracy?
3. How would the approach differ if the data followed a linear rather than exponential trend?
4. Can we use this function to predict values outside the provided range, and what are the limitations?
5. How would a logarithmic function fit if the growth wasn't strictly exponential?

Tip

When estimating exponential growth, checking the fit with $R^2$ helps ensure accuracy and reveals how well the model captures the trend.