Chapter 4.4, Problem 63E

a.

To determine

To describe: The graph of $y_4$ for $a=2,3,4,5$ and to generalize the description to an arbitrary $a>1$ .

Answer to Problem 63E

The graphs of $y_4=a$ for $a=2,3,4,5$ are horizontal lines parallel to $x=axis$ with $y-\mathrm{int}ercepts$ as $y=2,y=3,y=4,y=5$ respectively.

For arbitrary $a>1$ , the graph of $y_4=a$ is a straight line parallel to $x=axis$ with $\text{'}a\text{'}$ as the $y-\mathrm{int}ercept$ .

Explanation of Solution

Given Information: $y_1=a^x,y_2=NDER(y_1,x),y_3=\frac{y_2}{y_1},y_4=e^{y_3}$ .

Calculation:

  $y_1=a^x$

Differentiating both sides with respect to $x$ ,

  $\begin{array}{l}{y}_2=\frac{d{y}_1}{dx}\\ =a^x\ln a\end{array}$

  $\begin{array}{l}{y}_3=\frac{y_2}{y_1}\\ =\frac{a^x\ln a}{a^x}\\ =\ln a\end{array}$

  $\begin{array}{l}{y}_4=e^{y_3}\\ =e^{\ln a}\\ =a\end{array}$

Equation of a straight line is $y=mx+c$ , where $m$ is the slope and $c$ is the $y-\mathrm{int}ercept$ .

Comparing the equation $y_4=a$ with the above equation, it can be concluded that the graphs of $y_4=a$ for $a=2,3,4,5$ are horizontal lines parallel to $x=axis$ with $y-\mathrm{int}ercepts$ as $y=2,y=3,y=4,y=5$ respectively.

Graph:

  

For arbitrary $a>1$ , the graph of $y_4=a$ is a straight line parallel to $x=axis$ with $\text{'}a\text{'}$ as the $y-\mathrm{int}ercept$ .

b.

To determine

To describe: The graph of $y_3$ for $a=2,3,4,5$ and to generalize the description to an arbitrary $a>1$ . Also, to compare a table of values for $y_3$ for $a=2,3,4,5$ .

Answer to Problem 63E

The graphs of $y_3=\ln a$ for $a=2,3,4,5$ are horizontal lines parallel to $x=axis$ with $y-\mathrm{int}ercepts$ as $y=2,y=3,y=4,y=5$ respectively.

For arbitrary $a>1$ , the graph of $y_3=\ln a$ is a straight line parallel to $x=axis$ with $\text{'}a\text{'}$ as the $y-\mathrm{int}ercept$ .

Explanation of Solution

Given Information: $y_1=a^x,y_2=NDER(y_1,x),y_3=\frac{y_2}{y_1},y_4=e^{y_3}$ .

Calculation:

From part (a), $y_3=\ln a$ .

For $a=2,$

  $y_3=\ln 2=0.69314$

For $a=3,$

  $y_3=\ln 3=1.09861$

For $a=4,$

  $y_3=\ln 4=1.368295$

For $a=5,$

  $y_3=\ln 5=1.60944$

Equation of a straight line is $y=mx+c$ , where $m$ is the slope and $c$ is the $y-\mathrm{int}ercept$ .

Comparing the equation $y_3=\ln a$ with the above equation, it can be concluded that the graphs of $y_3=\ln a$ for $a=2,3,4,5$ are horizontal lines parallel to $x=axis$ with $y-\mathrm{int}ercepts$ as $y=2,y=3,y=4,y=5$ respectively.

Graph:

  

For arbitrary $a>1$ , the graph of $y_3=\ln a$ is a straight line parallel to $x=axis$ with $\text{'}a\text{'}$ as the $y-\mathrm{int}ercept$ .

c.

To determine

To explain: $\frac{d}{dx}{a}^x=a^{x}iffa=e$ .

Explanation of Solution

Given Information: $y_1=a^x,y_2=NDER(y_1,x),y_3=\frac{y_2}{y_1},y_4=e^{y_3}$ and results from part (a) and (b).

Calculation:

From part (a),

  $y_1=a^x$

  $\begin{array}{l}{y}_2=\frac{d{y}_1}{dx}\\ =\frac{d{a}^x}{dx}\\ =a^x\ln a\end{array}$

So, $\frac{d}{dx}{a}^x=a^x$

  $\begin{array}{l}\iff a^x\ln a=a^x\\ \iff \ln a=1\\ \iff a=e\end{array}$

Hence proved.

d.

To determine

To prove: $y_1=y_2$ if and only if $a=e$ .

Explanation of Solution

Given Information: $y_1=a^x,y_2=NDER(y_1,x),y_3=\frac{y_2}{y_1},y_4=e^{y_3}$ .

Proof:

  $y_1=a^x$

  $y_2=a^x\ln a$

Now, $y_1=y_2$

  $\begin{array}{l}\iff a^x\ln a=a^x\\ \iff \ln a=1\\ \iff a=e\end{array}$

Hence proved.