Chapter 4.7, Problem 14E

Solution

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

  $y=0.212\cdot {\left(4.88\right)}^x$ .

Given: points are given in $\left(x,y\right)$ as $\left(1,1.4\right)$ , $\left(2,6.7\right)$ , $\left(3,32.9\right)$ , $\left(4,161.4\right)$ and $\left(5,790.9\right)$ .

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 exponential model or function is defined as:

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

(3) 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:

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

  $\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ \ln y& 0.34& 1.91& 3.49& 5.08& 6.67\end{array}$

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

  

Now, the general exponential form is $y=a{b}^x$ ;

Here, choose two points $\left(1,0.34\right)$ and $\left(5,6.67\right)$ 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;

  $\begin{array}{c}\ln y-0.34=\frac{6.67-0.34}{5-1}\left(x-1\right)\\ \ln y-0.34=\frac{6.33}{4}\left(x-1\right)\\ \ln y-0.34=1.585\left(x-1\right)\\ \ln y-0.34=1.585x-1.585\end{array}$

Thus,

  $\begin{array}{c}\ln y=1.585x-1.585+0.34\\ \ln y=1.585x-1.551\end{array}$

Now, take the exponentiation of each side by e;

  $\begin{array}{c}{e}^{\ln y}=e^{1.585x-1.551}\left[e^{\ln y}=y\right]\\ y=e^{-1.551}\cdot e^{1.585x}\\ y\approx 0.212\cdot {\left[{\left(2.7182\right)}^{1.585}\right]}^x\left[\begin{array}{c}e\approx 2.7182\\ e^{1.585}\approx 4.879\\ e^{-1.551}\approx 0.212\end{array}\right]\\ y\approx 0.212\cdot {\left(4.879\right)}^x\end{array}$

Therefore, the exponential function of the model is $y=0.212\cdot {\left(4.88\right)}^x$ .

Conclusion:

Hence, the exponential function of the model that satisfies the given points is $y=0.212\cdot {\left(4.88\right)}^x$ .