Chapter 4.3, Problem 83E

a.

To determine

To define: to draw the graphs of the functions of family of , $f\left(x\right)={\log }_2\left({\log }_{10}x\right)$ .

Answer to Problem 83E

Domain : $\left(1,\infty \right),\{x|x\succ 1\}$

Range: $\left(-\infty ,\infty \right)\{y|y\in R\}$ .

  

Explanation of Solution

Given information:

The function to draw the graphs family of c=1,2,3,4 , $f\left(x\right)=\log \left(cx\right)$ .

The function todraw the graphs of the family of c=1,2,3,4 , $f\left(x\right)=\log \left(cx\right)$ .

As take x=1,

  $\begin{array}{c}y=\log \left(c\right)\\ x=2\\ y=\log \left(2c\right)\end{array}$

Now for x=3:

  $y=\log \left(3c\right)$ ,

The log function can be graphed using the vertical asymptote at x=0 , and the points $\left(1,\log \left(c\right)\right),\left(2,\log 2c\right),\left(3,\log 3c\right)$ .

The graph of the given equation:

  

b.

To determine

To define: relation of graph part ‘a’ of the function $f\left(x\right)=\log \left(cx\right)$ .

Answer to Problem 83E

The graph of part ‘a’ of the give function $f\left(x\right)=\log \left(cx\right)$ .

  

The graph of the function $f\left(x\right)=\log \left(cx\right)$ is th graph of $f\left(x\right)=\log \left(x\right)$ shifted upward log c units.

Explanation of Solution

Relation of graph part ‘a’ of the function $f\left(x\right)=\log \left(cx\right)$ .

Given information:

Relation of graph part ‘a’ of the function $f\left(x\right)=\log \left(cx\right)$ .

The graph of part ‘a’ of the give function $f\left(x\right)=\log \left(cx\right)$ .

  

The graph of the function $f\left(x\right)=\log \left(cx\right)$ is th graph of $f\left(x\right)=\log \left(x\right)$ shifted upward log c units.

The values of the function for every x,y, as x increases then y also increases.

The solution of the give equation