Chapter 4.7, Problem 35PS

Solution

To Graph: a scatter plot of given points and also find the exponential function of the given point.

  $y=x^{1.5}$ .

Given: points are given in $\left(x,y\right)$ as $\left(10,10\right)$ , $\left(20,12\right)$ , $\left(30,15\right)$ , $\left(40,25\right)$ , $\left(50,40\right)$ and $\left(60,100\right)$ .

Concept used:

(1) If a function $y=f\left(x\right)$ passes through a point $\left(x_0,y_0\right)$ , then the point $\left(x_0,y_0\right)$ satisfies the equation of the function, that is, $y_0=f\left(x_0\right)$ .

(2) A power model or function is defined as:

  $y=a\cdot {(x)}^b$ ; where $a\ne 0$ and $b\ge 0$ .

(4) An exponential model or function is as follows:

  $y=a\cdot {(b)}^x$ ; where $a\ne 0$ and $b>0$ .

(5) Equation of a line passing through the given two points as like $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ .

Then, $\left(y-y_1\right)=m\left(x-x_1\right)$ .

Here, $m$ is the slope. And $m$ is as follows:

  $m=\frac{y_2-y_1}{x_2-x_1}$ .

Calculation:

(a)The table of data pairs $\left(x,\ln y\right)$ written as;

  $\begin{array}{ccccccc}x& 10& 20& 30& 40& 50& 60\\ \ln y& 2.302& 2.485& 2.708& 3.219& 3.689& 4.605\end{array}$

The Scatter graph of $\left(x,\ln y\right)$ is given as;

  

Here, choose two points $\left(10,2.302\right)$ and $\left(60,4.605\right)$ then the equation of the line;

  $\ln y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

Now, substitute the value;

  $\begin{array}{c}\ln y-60=\frac{4.605-2.302}{60-10}\left(x-10\right)\\ \ln y+60=\frac{2.303}{50}\left(x-10\right)\\ \ln y+60=46.06\left(x-10\right)\\ \ln y+60=46.06x-460.6\end{array}$

Thus,

  $\begin{array}{c}\ln y=46.06x-460.6-60\\ \ln y=46.06x-520.6\end{array}$

Now, take the exponentiation of each side by e;

  $\begin{array}{c}{e}^{\ln \left(y\right)}=e^{46.06x-520.6}\left[e^{\ln a}=a\right]\\ y=e^{46.06x}\cdot e^{-520.6}\\ y\approx {\left[{\left(e\right)}^{46.06}\right]}^x\cdot 8.1865\left[\begin{array}{c}e\approx 2.7182\\ e^{-520.6}\approx 8.1865\\ e^{46.06}\approx 1.0069\end{array}\right]\\ y\approx 14920\cdot {\left(1.0069\right)}^x\end{array}$

Therefore, the exponential function of the model is $y=14920\cdot {\left(1.0069\right)}^x$ .

(b)The table of data pairs $\left(\ln x,\ln y\right)$ written as;

  $\begin{array}{ccccccc}\ln x& 2.302& 2.995& 3.402& 3.689& 3.912& 4.094\\ \ln y& 2.302& 2.485& 2.708& 3.219& 3.689& 4.605\end{array}$

The Scatter graph of $\left(\ln x,\ln y\right)$ is given as;

  

Here, choose two points $\left(2.302,2.302\right)$ and $\left(4.094,4.605\right)$ then the equation of the line;

  $\ln y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(\ln x-x_1\right)$

Now, substitute the value;

  $\begin{array}{c}\ln y-2.302=\frac{4.605-2.302}{4.094-2.302}\left(\ln x-2.302\right)\\ \ln y-2.302=\frac{2.303}{1.792}\left(\ln x-2.302\right)\\ \ln y-2.302=1.2851\cdot \left(\ln x-2.302\right)\\ \ln y-2.302=1.2851\ln x-2.9583\end{array}$

Thus,

  $\begin{array}{c}\ln y-1.2851\ln x=-2.9583+2.302\\ \ln y-\ln x^{1.2851}=-0.6563\left[a\ln b=\ln b^a\right]\\ \ln \left(\frac{y}{x^{1.2851}}\right)=-0.6563\end{array}$

Now, take the exponentiation of each side by e;

  $\begin{array}{c}{e}^{\ln \left(\frac{y}{x^{1.2851}}\right)}=e^{-0.6563}\left[e^{\ln a}=a\right]\\ \frac{y}{x^{1.2851}}=e^{-0.6563}\\ y\approx e^{-0.6563}\cdot x^{1.2851}\left[\begin{array}{l}e\approx 2.7182\\ e^{-0.6563}\approx 0.5187\end{array}\right]\\ y\approx 0.5187\cdot x^{1.2851}\end{array}$

Therefore, the power function of the model is $y=0.518\cdot x^{1.285}$

(C) By the graph of an exponential function of the data pairs represents the straight line and generates the scatter plot. But the power function of the data pairs is not representing the scatter plot. So, the power function is not the original function.

Then also it is clear the exponential function of the model is perfect showing these points in a scatter plot.

(D)

An exponential model or function is as follows:

  $y=a\cdot {(b)}^x$ ; where $a\ne 0$ and $b>0$ .

Therefore, the exponential function of the model is $y=14920\cdot {\left(1.0069\right)}^x$ .

Conclusion:

Hence, the exponential function of the model satisfies the given points be

  $y=14920\cdot {\left(1.0069\right)}^x$ and also represents the data pairs in the scatter plot. And the power function of the model be $y=0.518\cdot x^{1.285}$ and not satisfies the data pairs in the scatter plot.