Chapter 4.4, Problem 65E

(a)

To determine

To calculate: The equation for a line that is tangent to graph $y=\ln x$ passing through origin.

Answer to Problem 65E

The equation of line is $y=\frac{x}{e}$ .

Explanation of Solution

Given information: The equation of y in terms of x is given as,

  $y=\ln x$

Formula Used:

  $(y-y_1)=\frac{dy}{dx}(x-x_1)$

Calculation: Let us assume that the tangent at $(0,0)$ passes through origin,

  $\begin{array}{l}\text{Slope of }y=\ln x\text{ at }x=0\text{ is }{\frac{dy}{dx}|}_{x=0}=\frac{1}{e}\\ \text{So Equation of Tangent is,}\\ (y-0)=\frac{1}{e}(x-0)\\ ey=x\Rightarrow y=\frac{x}{e}\end{array}$

(b)

To determine

To compare: The curve $y=\ln x\text{ and line }y=\frac{x}{e}\text{ for all positive }x\ne e.$

Answer to Problem 65E

  $\text{Therefore, }\ln x<\frac{x}{e}\text{ for all positive }x\ne e.$

Explanation of Solution

Given information: The equation of y in terms of x is given as,

  $y=\ln x\text{ and }y=\frac{x}{e}$

The domain for $y=\ln x$ is given as,

  $\text{Domain:}\left\{x\in R:x>0\right\}$

  $\begin{array}{l}\text{The graph of }y=\frac{x}{e}\text{ has the domain as,}\\ \text{Domain:}\left\{x\in R:x\ne e\right\}\end{array}$

  $\text{Therefore, }\ln x<\frac{x}{e}\text{ for all positive }x\ne e.$

(c)

To determine

To prove: $\ln {(x)}^e\text{ <}x\text{ for all positive }x\ne e.$

Answer to Problem 65E

  $\text{Therefore, }\ln {(x)}^e<x\text{ for all positive }x\ne e.$

Explanation of Solution

Given information: The equation of y in terms of x is given as,

  $y=\ln {(x)}^e\text{ and }y=x$

The domain for $y=\ln {(x)}^e$ is given as,

  $\text{Domain:}\left\{x\in R:x>0\right\}$

  $\begin{array}{l}\text{The graph of }y=x\text{ has the domain as,}\\ \text{Domain:}\left\{x\in R:x\ne e\right\}\end{array}$

  

  $\text{Therefore, }\ln (x)<x\text{ for all positive }x\ne e.$

(d)

To determine

To prove: $x^e{\text{ <e}}^x\text{ for all positive }x\ne e.$

Answer to Problem 65E

  $\text{Therefore, }{x}^e<e^x\text{ for all positive }x\ne e.$

Explanation of Solution

Given information: The equation of y in terms of x is given as,

  $y=x^e\text{ and }y=e^x$

The domain for $y=x^e$ is given as,

  $\text{Domain:}\left\{x\in R:x\ge 0\right\}$

  $\begin{array}{l}\text{The graph of }y=e^x\text{ has the domain as,}\\ \text{Domain:}\left\{x\in R:x\ne e\right\}\end{array}$

  

  $\text{Therefore, }{x}^e<e^x\text{ for all positive }x\ne e.$

(e)

To determine

To prove: ${\pi }^e\text{ or }{e}^{\pi }\text{ which is bigger}\text{.}$

Answer to Problem 65E

  ${\text{Therefore, e}}^{\pi }>{\pi }^e.$

Explanation of Solution

Given information: The equation of y in terms of x is given as,

  $y={\pi }^e\text{ and }y=e^{\pi }$

The domain for $y={\pi }^e$ is given as,

  $\text{Domain:}\left\{x\in R\right\}$

  $\begin{array}{l}\text{The graph of }y=e^{\pi }\text{ has the domain as,}\\ \text{Domain:}\left\{x\in R\right\}\end{array}$

  

  ${\text{Therefore, e}}^{\pi }>{\pi }^e.$