Chapter 1.5, Problem 58E

(a)

To determine

To plot: The graph of functions $y_1=\ln \left(\frac{x}{a}\right)$ and $y_2=\ln \left(x\right)$ for different values of a . Also, explain the relation between $y_1$ and $y_2$ .

Explanation of Solution

Given information: The functions are $y_1=\ln \left(\frac{x}{a}\right)$ and $y_2=\ln \left(x\right)$ for $a=2,3,4,5$ .

Graph:

The function $y_1=\ln \left(\frac{x}{a}\right)$ for $a=2$ is $y_1=\ln \left(\frac{x}{2}\right)$ , for $a=3$ is $y_1=\ln \left(\frac{x}{3}\right)$ , for $a=4$ is $y_1=\ln \left(\frac{x}{4}\right)$ and for $a=5$ is $y_1=\ln \left(\frac{x}{5}\right)$ .

Draw the graph of the functions $y_1=\ln \left(\frac{x}{a}\right)$ and $y_2=\ln \left(x\right)$ as shown below.

  

Figure (1)

Interpretation: From the above graph it can be observed that the graph of function $y_2=\ln \left(x\right)$ is a vertical shift of $y_1=\ln \left(\frac{x}{a}\right)$ . Also, it can be interpreted that the graph of $y_2$ can be drawn for $a=1$ from the function $y_1=\ln \left(\frac{x}{a}\right)$ .

(b)

To determine

To plot: The graph of functions $y_3=y_2-y_1$ for different values of a .

Explanation of Solution

Given information: The functions are $y_1=\ln \left(\frac{x}{a}\right)$ , $y_2=\ln \left(x\right)$ and $y_3=y_2-y_1$ for $a=2,3,4,5$ .

Graph:

For $a=2$ the function is $y_3=\ln x-\ln \left(\frac{x}{2}\right)$ .

For $a=3$ the function is $y_3=\ln x-\ln \left(\frac{x}{3}\right)$ .

For $a=4$ the function is $y_3=\ln x-\ln \left(\frac{x}{4}\right)$ .

For $a=5$ the function is $y_3=\ln x-\ln \left(\frac{x}{5}\right)$ .

Draw the graph of the functions $y_1=\ln \left(\frac{x}{a}\right)$ and $y_2=\ln \left(x\right)$ as shown below.

  

Figure (2)

Interpretation: From the above figure it can be observed that the graph of $y_3=\ln x-\ln \left(\frac{x}{a}\right)$ for different values of a is a horizontal line.

(c)

To determine

To plot: The graph of functions $y_4=e^{y_3}$ for different values of a . Also, compare the graphs with the graph of $y=a$ .

Explanation of Solution

Given information: The functions are $y_1=\ln \left(\frac{x}{a}\right)$ , $y_2=\ln \left(x\right)$ , $y_3=y_2-y_1$ and $y_4=e^{y_3}$ for $a=2,3,4,5$ .

Graph:

The functions $y_3$ for $a=2,3,4,5$ are $y_3=\ln x-\ln \left(\frac{x}{2}\right)$ , $y_3=\ln x-\ln \left(\frac{x}{3}\right)$ , $y_3=\ln x-\ln \left(\frac{x}{4}\right)$ and $y_3=\ln x-\ln \left(\frac{x}{5}\right)$ respectively.

Consider that $y_4=e^{y_3}$ . So, the functions $y_4$ for $a=2,3,4,5$ are $y_4=e^{\ln x-\ln \left(\frac{x}{2}\right)}$ , $y_4=e^{\ln x-\ln \left(\frac{x}{3}\right)}$ , $y_4=e^{\ln x-\ln \left(\frac{x}{4}\right)}$ and $y_4=e^{\ln x-\ln \left(\frac{x}{5}\right)}$ respectively.

The equations of $y=a$ for $a=2,3,4,5$ are $y=2$ , $y=3$ , $y=4$ and $y=5$ respectively.

Draw the graph of the functions $y_4=e^{y_3}$ and $y=a$ as shown below.

  

Figure (3)

Interpretation: From the above figure it can be observed that the graph of $y_4=e^{y_3}$ for different values of a is the same as the graph of $y=a$ .

(d)

To determine

To find: The solution for $y_1$ by using $e^{y_3}=e^{y_2-y_1}=a$ .

Answer to Problem 58E

The solution for $y_1$ is $y_1=\ln x-\ln a$ .

Explanation of Solution

Given information: The functions are $y_1=\ln \left(\frac{x}{a}\right)$ , $y_2=\ln \left(x\right)$ , $y_3=y_2-y_1$ and $y_4=e^{y_3}$ for $a=2,3,4,5$ . The equation is $e^{y_3}=e^{y_2-y_1}=a$ .

Calculation:

Take $\ln$ of both sides of the equation $e^{y_2-y_1}=a$ .

  $\begin{array}{c}\ln \left(e^{y_2-y_1}\right)=\ln \left(a\right)\\ y_2-y_1=\ln \left(a\right)\\ y_1=y_2-\ln a\\ y_1=\ln x-\ln a\end{array}$

Therefore, the solution for $y_1$ is $y_1=\ln x-\ln a$ .