To perform exponential regression on the dataset, we need to find an exponential model in the form:

$y = a \cdot e^{(b \cdot x)}$

where:

• $y$ is the output (dependent variable),
• $x$ is the input (independent variable),
• $a$ and $b$ are constants to be determined.

The given data points are:

$x: 1, 2, 3, 4, 5$ $y: 53.0, 55.2, 57.4, 59.7, 62.0$

Step 1: Transform the Exponential Equation to a Linear Form

Since the exponential function $y = a \cdot e^{(b \cdot x)}$ is not linear, we can take the natural logarithm of both sides to linearize it:

$\ln(y) = \ln(a \cdot e^{(b \cdot x)}) = \ln(a) + b \cdot x$

Let:

• $Y = \ln(y)$ (transformed dependent variable),
• $A = \ln(a)$,
• $b$ remains the slope.

This transformation gives a linear equation: $Y = A + b \cdot x$

Now we can use linear regression to find the values of $A$ and $b$, and then determine $a = e^A$.

Step 2: Transform the Data

Let's calculate $\ln(y)$ for each $y$ value.

$y: 53.0, 55.2, 57.4, 59.7, 62.0$ $\ln(y): \ln(53.0), \ln(55.2), \ln(57.4), \ln(59.7), \ln(62.0)$

Computing each:

$\ln(53.0) \approx 3.9703$ $\ln(55.2) \approx 4.0100$ $\ln(57.4) \approx 4.0492$ $\ln(59.7) \approx 4.0884$ $\ln(62.0) \approx 4.1271$

Thus, we have the transformed data set:

$x: 1, 2, 3, 4, 5$ $Y: 3.9703, 4.0100, 4.0492, 4.0884, 4.1271$

Step 3: Calculate $b$ and $A$ Using Linear Regression Formulas

To find $b$ (slope) and $A$ (intercept), we use the formulas:

$b = \frac{n \sum(xY) - \sum x \sum Y}{n \sum(x^2) - (\sum x)^2}$ $A = \frac{\sum Y - b \sum x}{n}$

where $n = 5$ (the number of data points).

Let's calculate each component:

1. Sum of $x$: $\sum x = 1 + 2 + 3 + 4 + 5 = 15$

2. Sum of $Y$: $\sum Y = 3.9703 + 4.0100 + 4.0492 + 4.0884 + 4.1271 = 20.245$

3. Sum of $x^2$: $\sum(x^2) = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$

4. Sum of $x \cdot Y$: $\sum(xY) = (1 \cdot 3.9703) + (2 \cdot 4.0100) + (3 \cdot 4.0492) + (4 \cdot 4.0884) + (5 \cdot 4.1271) = 84.92$

Now we can plug these values into the formulas.

Calculating $b$:

$b = \frac{5 \cdot 84.92 - 15 \cdot 20.245}{5 \cdot 55 - 15^2}$ $b = \frac{424.6 - 303.675}{275 - 225} = \frac{120.925}{50} = 2.4185$

Calculating $A$:

$A = \frac{20.245 - (2.4185 \cdot 15)}{5}$ $A = \frac{20.245 - 36.2775}{5} = \frac{-16.0325}{5} = -3.2065$

Step 4: Determine $a$

Since $A = \ln(a)$, we have: $a = e^{-3.2065} \approx 0.0406$

Step 5: Write the Exponential Model

With $a \approx 0.0406$ and $b \approx 2.4185$, our exponential model is:

$y = 0.0406 \cdot e^{(2.4185 \cdot x)}$

This model represents the best exponential fit for the given data.

Would you like further clarification on any of the steps?

Here are five questions related to exponential regression and this model:

1. How does changing the base of the exponential function affect the model?
2. Why is taking the natural logarithm useful for exponential regression?
3. How would you interpret the constant $b$ in this context?
4. Can exponential regression be applied to data that decreases over time?
5. What are some real-world applications of exponential regression?

Tip: Remember, exponential regression is particularly useful for data that grows (or decays) at an increasing rate, such as population growth or radioactive decay.