Chapter 3.4, Problem 78E

To determine

To find : solution to the equation algebraically and verify the result graphically.

Answer to Problem 78E

The solution to the equation $\left(1-\ln x\right)/x^2=0$ are $\underset{\_}{\boxed{x=e}}$

Explanation of Solution

Given information: $\frac{1-\ln x}{x^2}=0$

Concept Involved:

The word “solve” means the process of finding the value of x that makes the equation true.

In order to solve we need to undo whatever is done to x .

The solution to the equation $f\left(x\right)=g\left(x\right)$ graphically is the x value of the point of intersection of where the two graphs meet.

Formula Used:

1. If logarithmic equation is $\ln A=m$ , then the equivalent exponential equation is $A=e^m$
2. $\ln e=1$
3. Zero factor property: If $a\cdot b=0$ , then $a=0\left(or\right)b=0$

Calculation:

Rewrite the left side of the equation using the negative exponent rule $\frac{1}{a^m}=a^{-m}$

  $x^{-2}\left(1-\ln x\right)=0$

Use the zero factor property to set each factor to zero

  $x^{-2}=0$▶ 1^st equation

  $1-\ln x=0$▶2^nd equation

Solving the equation $x^{-2}=0$

• There is no value of x that makes the above equation TRUE, so No solution

Solving the equation $1-\ln x=0$

• By subtracting 1 on both sides, then
• By Simplifying on both sides, then
• By dividing -1 on both sides, then
• By simplifying fraction on both sides of the equation, and then
• Rewrite the logarithmic equation to exponential equation
$1-\ln x-1=0-1$

  $-\ln x=-1$

  $\frac{-\ln x}{-1}=\frac{-1}{-1}$

  $\ln x=1$

  $x=e^1$

Check for extraneous solution by substituting the result in the original equation.

  $x=e$ makes the original equation TRUE.

Graph:

  

Interpretation:

Setting up left side of the equation $f\left(x\right)=\left(1-\ln x\right)/x^2$ and right side of the equation $g\left(x\right)=0$ .

The xcoordinate of point of intersection of two graphs is the solution to the equation $\left(1-\ln x\right)/x^2=0$

Conclusion:

The solution to the equation $\left(1-\ln x\right)/x^2=0$ are $\underset{\_}{\boxed{x=e}}$