Exercise 25 Page 347

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

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

(30, 2.71)

and

(60, 4.61)

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 $(30, 2.71)$ & $(60, 4.61)$

m = \frac{4 . 6 1 - 2 . 7 1}{6 0 - 3 0}

Subtract terms

m = \frac{1 . 9}{3 0}

Calculate quotient

m = 0.063333 \dots

Round to $2$ decimal place(s)

m \approx 0.06

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

(30, 2.71)

be our reference point.

ln y - 2.71 = 0.06 (x - 30)

Next, we will isolate

y

to have the model in the form of

y = a b^{x} .

ln y - 2.71 = 0.06 (x - 30)

▼

Solve for

y

Distribute $0.06$

ln y - 2.71 = 0.06 x - 1.8

$LHS + 2.71 = RHS + 2.71$

ln y = 0.06 x + 0.91

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

e^{l n y} = e^{0.06 x + 0.91}

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

y = e^{0.06 x + 0.91}

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

y = e^{0.06 x} \cdot e^{0.91}

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

y = (e^{0.06})^{x} \cdot e^{0.91}

Calculate power

y = (1.06183 \dots)^{x} \cdot 2.48432 \dots

y = 2.48432 \dots (1.06183 \dots)^{x}

Round to $2$ decimal place(s)

y = 2.48 (1.06)^{x}

As a result, one possible model is

y = 2.48 (1.06)^{x}

depending on the choice of the points.