Chapter 3.4, Problem 71E

Solution

To prove : The change of base formula for logarithms.

Given information :

The change of base formula for logarithms.

Proof :

Consider the change of base formula for logarithms,

  ${\log }_{b}x=\frac{{\log }_{a}x}{{\log }_{a}b}$ for all positive real numbers $a,b,\text{ and }x$ , where $a\ne 1$ .

Let

  $\begin{array}{c}{\log }_{b}x=y\\ b^y=x\\ {\log }_a{b}^y={\log }_{a}x\\ y{\log }_{a}b={\log }_{a}x\\ y=\frac{{\log }_{a}x}{{\log }_{a}b}\\ {\log }_{b}x=\frac{{\log }_{a}x}{{\log }_{a}b}\end{array}$

Therefore, it is proved that ${\log }_{b}x=\frac{{\log }_{a}x}{{\log }_{a}b}$ .

(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 $x>-3$ .