Chapter 4.7, Problem 33PS

Solution

i.

To graph: The scatter plot of $\left(x,\ln y\right)$ data pairs and find the exponential model.

  $y=0.5{\left(1.99\right)}^x$

Given:

The data pairs $\left(1,1\right),\left(2,2\right),\left(3,5\right),\left(4,7\right),\left(5,20\right),\left(6,40\right),\left(7,80\right)$.

Concept Used:

1. To get the scatter plot first create the $\left(x,\ln y\right)$ table and then plot the coordinates on the graph using the graphing calculator.
2. Use the point-slope formula $y-y_1=m\left(x-x_1\right)$ to determine the equation of a line.
3. Use the formula$m=\frac{y_2-y_1}{x_2-x_1}$ to compute the slope where $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are coordinates.

Calculation:

Use the calculator to create the table of $\left(x,\ln y\right)$ data pairs.

Years $\left(x\right)$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
Business users $\left(\ln y\right)$	$0$	$0.693$	$1.609$	$1.946$	$2.996$	$3.689$	$4.382$

The following graph has been obtained using the above data pairs:

  

Let $\left(x_1,y_1\right)=\left(1,0\right)$ and $\left(x_2,y_2\right)=\left(2,0.693\right)$ . Substitute the coordinates in point slope equation to obtain the equation and then solve for $y$ to get the exponential equation.

  $\begin{array}{c}\ln y-0=\left(\frac{0.693-0}{2-1}\right)\left(x-1\right)\\ \ln y-0=0.693x-0.693\\ \ln y=0.693x-0.693\end{array}$

Solve for $y$.

  $\begin{array}{l}y=e^{-0.693}{\left(e^{0.693}\right)}^x\\ y=0.5{\left(1.99\right)}^x\end{array}$

Conclusion:

The exponential model is $y=0.5{\left(1.99\right)}^x$.

ii.

To explain: Which model the describes the data better and find the best model.

The power model describes the non-business users better and the power model is $y=54.981x^{0.8196}$.

Given:

The data pairs $\left(1,55\right),\left(2,97\right),\left(3,140\right),\left(4,160\right),\left(5,195\right),\left(6,235\right),\left(7,260\right)$.

Concept Used:

1. The models whose coordinates lies close to the line is said to be the best model.
2. Use the point-slope formula $y-y_1=m\left(x-x_1\right)$ to determine the equation of a line.
3. Use the formula$m=\frac{y_2-y_1}{x_2-x_1}$ to compute the slope where $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are coordinates.

Calculation:

Use the graphing calculator to graph the given coordinates of $\left(x,y\right)$ data pair to check that linear model is best or not.

  

Similarly, obtain the graph of $\left(x,\ln y\right)$ data pairs to check that the exponential model is best or not.

  

And finally obtain the graph of $\left(\ln x,\ln y\right)$ data pairs using the graphing calculator to check that power model is best or not.

  

The coordinates of the $\left(\ln x,\ln y\right)$ data pairs fits the line better compared to other two graphs. So, it can be concluded that the power function best models the number of non-business users over time.

To find the power model, consider $\left(x_1,y_1\right)=\left(0,4.007\right)$ and $\left(x_2,y_2\right)=\left(0.693,4.575\right)$ . Substitute the coordinates in the point slope equation.

  $\begin{array}{l}\ln y-4.007=\left(\frac{4.575-4.007}{0.693-0}\right)\left(\ln x-0\right)\\ \ln y-4.007=0.8196\ln x\\ \ln y=\ln x^{0.8196}+4.007\end{array}$

Solve for $y$.

  $\begin{array}{l}y=e^{4.007}{e}^{\ln x^{0.8196}}\\ y=54.981x^{0.8196}\end{array}$

Conclusion:

The power function is best model because the coordinates of $\left(\ln x,\ln y\right)$ data pairs lies close to the line. The power model is $y=54.981x^{0.8196}$.