Chapter 6.7, Problem 26E

To determine

To show : An exponential model should be the good fit for the original data pairs $\left(x,y\right)$ and then write an exponential model for the original data.

Answer to Problem 26E

The exponential function that models the data is $y=\left(6.4\right){\left(1.4\right)}^x$ .

Explanation of Solution

Given information :

The given data is $\begin{array}{cccccccc}x& 1& 2& 3& 4& 5& 6& 7\\ y& 9& 14& 19& 25& 37& 53& 71\end{array}$ .

Calculation :

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

  $\begin{array}{cccccccc}x& 1& 2& 3& 4& 5& 6& 7\\ y& 2.2& 2.64& 2.94& 3.22& 3.61& 3.97& 4.26\end{array}$

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 $\left(1,2.2\right)$ and $\left(7,\text{ 4}\text{.26}\right)$ . Use these points to write an equation of the line. Then solve for y.

Therefore,

  $\begin{array}{l}\ln y-2.2=\frac{4.26-2.2}{7-1}\left(x-1\right)\text{ (Equation of the line)}\\ \ln y-2.2=0.34\left(x-1\right)\text{ (Simplify) }\\ \ln y=0.34x+1.86\\ y=e^{0.34x+1.86}\text{ (Exponentiate each side using base e)}\\ y=e^{0.34x}\left(e^{1.86}\right)\text{ (Use peoperty of exponents)}\\ y=\left(6.4\right){\left(1.4\right)}^x\text{ (Simplify)}\end{array}$

Hence,

The exponential function that models the data is $y=\left(6.4\right){\left(1.4\right)}^x$ .