Chapter 6.3, Problem 66E

To determine

To calculate: The correct order from least to greatest average rate of change over the time interval $1\le x\le 10$ .

Answer to Problem 66E

The required order is $c<a<b<d$ .

Explanation of Solution

Given information:

The given functions are

  

Calculation:

Consider the logarithm expression a. $y={\log }_{6}x$ .

Substitute $x=1$ in the above function.

  $\begin{array}{c}y={\log }_{6}1\\ =0\end{array}$

Again substitute $x=10$ in the above function.

  $\begin{array}{c}y={\log }_{6}10\\ =1.28\end{array}$

Consider the logarithm expression b. $y={\log }_{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}x$ .

Substitute $x=1$ in the above function.

  $\begin{array}{c}y={\log }_{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}1\\ =0\end{array}$

Again substitute $x=10$ in the above function.

  $\begin{array}{c}y={\log }_{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}10\\ =-4.5\end{array}$

Again from the graph of the given function c.

At $x=1,y=0$ and at $x=10,y=0.8$ .

Similarly from the given graph of the function d.

At $x=1,y=0$ and at $x=10,y=8$ .

Hence the required order is $c<a<b<d$ .