We will first create a table of data pairs $(x, ln y)$ using the given table.

Next, we will plot the transformed points.

The points lie close to a line, so an exponential model should be a good fit for the original data. To write an exponential model

y = a b^{x},

let's use the points

(1, 1.19)

and

(4, 2.31)

because they appear to be on the line. With this, we will write an equation in point-slope form.

ln y - l n y_{1} = m (x - x_{1})

In this form,

m

is the slope and

(x_{1}, l n y_{1})

is a reference point that is on the line. Using the Slope Formula, we can substitute the chosen points and find the slope.

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Substitute $(1, 1.19)$ & $(4, 2.31)$

m = \frac{2 . 3 1 - 1 . 1 9}{4 - 1}

Subtract terms

m = \frac{1 . 1 2}{3}

Calculate quotient

m = 0.37333 \dots

Round to $2$ decimal place(s)

m \approx 0.37

Now that we found the slope, we can write the equation in point-slope form. Let the point

(1, 1.19)

be our reference point.

ln y - 1.19 = 0.37 (x - 1)

Next, we will isolate

y

to have the model in the form of

y = a b^{x} .

ln y - 1.19 = 0.37 (x - 1)

▼

Solve for

y

Distribute $0.37$

ln y - 1.19 = 0.37 x - 0.37

$LHS + 1.19 = RHS + 1.19$

ln y = 0.37 x + 0.82

$e^{LHS} = e^{RHS}$

e^{l n y} = e^{0.37 x + 0.82}

$a = e^{l n (a)}$

y = e^{0.37 x + 0.82}

$a^{m + n} = a^{m} \cdot a^{n}$

y = e^{0.37 x} \cdot e^{0.82}

$(a^{m})^{n} = a^{m \cdot n}$

y = (e^{0.37})^{x} \cdot e^{0.82}

Calculate power

y = (1.44773 \dots)^{x} \cdot 2.27049 \dots

y = 2.27049 \dots (1.44773 \dots)^{x}

Round to $2$ decimal place(s)

y = 2.27 (1.45)^{x}

As a result, one possible model is

y = 2.27 (1.45)^{x}

depending on the choice of the points.

Exercise 28 Page 347

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