Chapter 3.4, Problem 70E

Solution

(a)

To calculate : The domain of the given functions and compare.

The domain of $f\left(x\right)\text{ and }g\left(x\right)$ is $x>3$ .

Given information :

The functions: $f\left(x\right)=2\ln x+\ln \left(x-3\right)\text{ and }g\left(x\right)=\ln x^2\left(x-3\right)$ .

Explanation :

Consider the functions $f\left(x\right)=2\ln x+\ln \left(x-3\right)\text{ and }g\left(x\right)=\ln x^2\left(x-3\right)$

Graph both the functions.

  

Both the graphs overlap, this means $f\left(x\right)=g\left(x\right)$ . So, both the functions will have same domain and range.

Therefore, the domain of $f\left(x\right)\text{ and }g\left(x\right)$ is $x>3$ .

(b)

To calculate : The domain of the given functions and compare.

The domain of $f\left(x\right)$ is $x>5$ and the domain of $g\left(x\right)$ is $x<-5,x>5$ .

Given information :

The functions: $f\left(x\right)=\ln \left(x+5\right)-\ln \left(x-5\right)\text{ and }g\left(x\right)=\ln \frac{x+5}{x-5}$ .

Explanation :

Consider the functions $f\left(x\right)=\ln \left(x+5\right)-\ln \left(x-5\right)\text{ and }g\left(x\right)=\ln \frac{x+5}{x-5}$

Graph both the functions.

  

Both the graphs overlap for $x>5$ , this means $f\left(x\right)=g\left(x\right)$ for $x>5$ .

Therefore, the domain of $f\left(x\right)$ is $x>5$ and the domain of $g\left(x\right)$ is $x<-5,x>5$ .

(c)

To calculate : The domain of the given functions and compare.

The domain of $f\left(x\right)$ is $x\ne -3$ and the domain of $g\left(x\right)$ is $x>-3$ .

Given information :

The functions: $f\left(x\right)=\log {\left(x+3\right)}^2\text{ and }g\left(x\right)=2\log \left(x+3\right)$ .

Explanation :

Consider the functions $f\left(x\right)=\log {\left(x+3\right)}^2\text{ and }g\left(x\right)=2\log \left(x+3\right)$

Graph both the functions.

  

Both the graphs overlap for $x>-3$ , this means $f\left(x\right)=g\left(x\right)$ for $x>-3$ .

Therefore, the domain of $f\left(x\right)$ is $x\ne -3$ and the domain of $g\left(x\right)$ is $x>-3$ .