Chapter 3.4, Problem 79E

To determine

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

Answer to Problem 79E

The solution to the equation $2x\ln x+x=0$ are $\underset{\_}{\boxed{x=1/\sqrt{e}}}$

Explanation of Solution

Given information: $2x\ln x+x=0$

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. Zero factor property: If $a\cdot b=0$ , then $a=0\left(or\right)b=0$

Calculation:

Factor the GCF in left side of the equation

  $x\left(2\ln x+1\right)=0$

Use the zero factor property to set each factor to zero

  $x=0$ ▶1^st equation

  $2\ln x+1=0$▶2^nd equation

Solving the equation $2\ln x+1=0$

• By subtracting 1 on both sides, then
• By Simplifying on both sides, then
• By dividing 2 on both sides, then
• By simplifying fraction on both sides of the equation, and then
• Rewrite the logarithmic equation to exponential equation
$2\ln x+1-1=0-1$

  $2\ln x=-1$

  $\frac{2\ln x}{2}=\frac{-1}{2}$

  $\ln x=-\frac{1}{2}$

  $x=e^{-1/2}$

Use the exponent rule $a^{-m}=\frac{1}{a^m}$ to rewrite the equation

  $x=\frac{1}{e^{1/2}}$

Simplify the right side of the equation

  $x\approx 0.6065306597$

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

  $x=\frac{1}{\sqrt{e}}$ makes the original equation TRUE.

Graph:

  

Interpretation:

Setting up left side of the equation $f\left(x\right)=2x\ln x+x$ 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 $2x\ln x+x=0$

Conclusion:

The solution to the equation $2x\ln x+x=0$ are $\underset{\_}{\boxed{x=1/\sqrt{e}}}$