Chapter 6.7, Problem 29E

To determine

To create : A scatter plot of the points $\left(x,\text{ ln}y\right)$ to determine whether an exponential model fits the data, if so , find an exponential model for the data.

Answer to Problem 29E

The exponential function that models the data is $y=14.59{\left(1.032\right)}^x$ .

Explanation of Solution

Given information :

The given data is $\begin{array}{cccccc}x& -13& -6& 1& 8& 15\\ y& 9.8& 12.2& 15.2& 19& 23.8\end{array}$ .

Calculation :

Create a table of data pairs $\left(x,\text{ ln}y\right)$ .

  $\begin{array}{cccccc}x& -13& -6& 1& 8& 15\\ \ln y& 2.28& 2.5& 2.72& 2.94& 3.17\end{array}$

Plot the above transformed points as shown below:

  

Since, the point lie close to a line, so an exponential model should be a good fit for the original data.

Find an exponential model $y=a{b}^x$ by choosing any two points on the line, such as $\left(1,\text{ 2}\text{.72}\right)$ and $\left(15,\text{ 3}\text{.17}\right)$ . Use these points to write an equation of the line. Then solve for y.

Therefore,

  $\begin{array}{l}\ln y-2.72=\frac{3.17-2.72}{15-1}\left(x-1\right)\text{ (Equation of the line)}\\ \ln y-2.72=0.032\left(x-1\right)\text{ (Simplify) }\\ \ln y=0.032x+2.68\\ y=e^{0.032x+2.68}\text{ (Exponentiate each side using base e)}\\ y=e^{0.032x}\left(e^{2.68}\right)\text{ (Use peoperty of exponents)}\\ y=14.59{\left(1.032\right)}^x\text{ (Simplify)}\end{array}$

Hence,

The exponential function that models the data is $y=14.59{\left(1.032\right)}^x$ .