Chapter 3.4, Problem 69E

(a)

To determine

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

Answer to Problem 69E

Graph is given.

Explanation of Solution

Given information:

  $y_1=a^x,y_2=NDER\left(y_1,x\right),y_3=\frac{y_2}{y_1},y_4=e^{y_3}$

Calculation:

The meaning of $NDER\left(y_1,x\right)$ is to differentiate $y_1$ and find derivative in the form of $y_2=f\left(x\right)$ .

Differentiate $y_1=a^x$ with respect to $x$ :

  $\begin{array}{l}\frac{d{y}_1}{dx}=a^x\ln a\\ y_2=a^x\ln a\end{array}$

To find the value of $y_3=\frac{y_2}{y_1}$ substitute the value of $y_2$ and $y_1$ in $y_3=\frac{y_2}{y_1}$ .

  $\begin{array}{l}{y}_3=\frac{a^x\ln a}{a^x}\\ y_3=\ln a\end{array}$

Now given that $y_4=e^{y_3}$ substitute the value of $y_3$ :

  $\begin{array}{l}{y}_4=e^{\ln a}\\ y_4=a\end{array}$

Draw the graph of $y_4=a$ for $a=2,3,4,5$ by substuting these value in place of $a$ .

The graph will be the horizontal lines as $y_4=a$ is an equation of line which intersect at $y=a$ and has the zero slope.

When $a=2$

  $y_4=a$ becomes:

  $y_4=2$

When $a=3$

  $y_4=a$ becomes:

  $y_4=3$

Similarly, for other values of $a$ equation of lines will be $y_4=4$ , $y_4=5$

The graph for different values of $a$ is given below:

  

For equation $y_1=a^x$ if $a>1$

  $y_1=a^x$ is an exponential function if the graph is drawn of that function is below:

  

It can be seen that as the exponent increases, the curves get steeper and the rate of growth increases respectively. A function which grows faster than a polynomial function is $y=a^x$ where $a>1$ . Thus, for any of the positive integers $n$ the function $f\left(x\right)$ is said to grow faster than that of $f_n\left(x\right)$

Thus, the exponential function having base greater than 1, i.e., $\text{a > 1 }$ is defined as $y=a^x$ . The domain of exponential function will be the set of entire real numbers R and the range are said to be the set of all the positive real numbers.

It must be noted that the exponential function is increasing and the point $\text{(0, 1) }$ always lies on the graph of an exponential function. Also, it is very close to zero if the value of $x$ is mostly negative.

(b)

To determine

To find: The graph of $y_3$ for $a=2,3,4,5$ .Compare the table of values for $y_3$and $\ln a$ .

Answer to Problem 69E

The graph is given and values of $y_3$and $\ln a$ are equal. the values of $y_3=\ln a$ can only be calculated when $a>1$ .

Explanation of Solution

Given information:

  $y_1=a^x,y_2=NDER\left(y_1,x\right),y_3=\frac{y_2}{y_1},y_4=e^{y_3}$

Calculation:

The meaning of $NDER\left(y_1,x\right)$ is to differentiate $y_1$ and find derivative in the form of $y_2=f\left(x\right)$ .

Differentiate $y_1=a^x$ with respect to $x$ :

  $\begin{array}{l}\frac{d{y}_1}{dx}=a^x\ln a\\ y_2=a^x\ln a\end{array}$

To find the value of $y_3=\frac{y_2}{y_1}$ substitute the value of $y_2$ and $y_1$ in $y_3=\frac{y_2}{y_1}$ .

  $\begin{array}{l}{y}_3=\frac{a^x\ln a}{a^x}\\ y_3=\ln a\end{array}$

Draw the graph of $y_3=\ln a$ for $a=2,3,4,5$ by substuting these value in place of $a$ .

When $a=2$

  $y_3=\ln 2$

When $a=3$

  $y_3=\ln 3$

When $a=4$

  $y_3=\ln 4$

The graph of $y_3=\ln a$ for $a=2,3,4,5$ is given below:

  

For $a=1$

  $y_3=\ln a$ will give the value $y_3=\ln \left(1\right)$ or $y_3=0$ .

Therefore the values of $y_3=\ln a$ can only be calculated when $a>1$ .

Basically the value found of $y_3$ is $\ln a$ so both values of $y_3$ and $\ln a$ represents the same values or it can be written as:

  $y_3=\frac{e^{x\ln a}\cdot \ln a}{a^x}$

Here, $a^x=e^{x\ln a}$

Table of values of $y_3$ for $a=2,3,4,5$ is given below:

$a$	2	3	4	5
$y=\ln a$	0.69	1.098	1.386	1.60

(c)

To determine

To find: How part $\left(a\right)$ and $\left(b\right)$ supports the statement $\frac{d}{dx}{a}^x=a^x$ if and only if $a=e$

Explanation of Solution

Given information:

  $y_1=a^x,y_2=NDER\left(y_1,x\right),y_3=\frac{y_2}{y_1},y_4=e^{y_3}$

Calculation:

The meaning of $NDER\left(y_1,x\right)$ is to differentiate $y_1$ and find derivative in the form of $y_2=f\left(x\right)$ .

Differentiate $y_1=a^x$ with respect to $x$ :

  $\begin{array}{l}\frac{d{y}_1}{dx}=a^x\ln a\\ y_2=a^x\ln a\end{array}$

To find the value of $y_3=\frac{y_2}{y_1}$ substitute the value of $y_2$ and $y_1$ in $y_3=\frac{y_2}{y_1}$ .

  $\begin{array}{l}{y}_3=\frac{a^x\ln a}{a^x}\\ y_3=\ln a\end{array}$

In part (a) the value $y_2$ is found $y_2=a^x\ln a$

  $a^x$ can be written as $e^{x\ln a}$ therefore $y_2$ becomes:

  $\begin{array}{l}\frac{d{y}_1}{dx}=e^{x\ln a}\ln a\\ or\\ \frac{d}{dx}\left(a^x\right)=e^{x\ln a}\ln a\end{array}$

If $a=e$ is substituted in $\frac{d}{dx}\left(a^x\right)=e^{x\ln a}\ln a$ it gives:

  $\begin{array}{l}\frac{d}{dx}\left(a^x\right)=e^{x\ln a}\ln (e)\\ \frac{d}{dx}\left(a^x\right)=e^{\ln a}{}^x\ln (e)\end{array}$

Recall the fact that exponent and log functions are inverse to each other therefore $\ln (e)=1$ .

Hence

  $\frac{d}{dx}\left(a^x\right)=a^x$

This proves that if $a=e$ then only $\frac{d}{dx}\left(a^x\right)=a^x$ is true.

Hence proved.

(d)

To determine

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

Explanation of Solution

Given information:

  $y_1=a^x,y_2=NDER\left(y_1,x\right),y_3=\frac{y_2}{y_1},y_4=e^{y_3}$

Calculation:

The meaning of $NDER\left(y_1,x\right)$ is to differentiate $y_1$ and find derivative in the form of $y_2=f\left(x\right)$ .

Differentiate $y_1=a^x$ with respect to $x$ :

  $\begin{array}{l}\frac{d{y}_1}{dx}=a^x\ln a\\ y_2=a^x\ln a\mathrm{....}(1)\end{array}$

  $a^x$ can be written as $e^{x\ln a}$ hence $y_1$ and $y_2$ becomes:

  $\begin{array}{l}{y}_1=a^x\ln (e)\mathrm{....}(2)\\ \ln (e)=1\end{array}$

If $a=e$ then equation $\left(2\right)$ becomes $y_1=a^x\ln (a)$ which is equal to equation $\left(1\right)$

  $y_2=e^{x\ln a}\ln a$

Therefore it can be written as $y_1=y_2$ if and only if $a=e$ .