Chapter 3.4, Problem 77E

To determine

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

Answer to Problem 77E

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

Explanation of Solution

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

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:

If logarithmic equation is $\ln A=m$ , then the equivalent exponential equation is $A=e^m$

  $\ln e=1$

Calculation:

Multiply 2 on both sides of the equation to get rid of the fraction

  $2\cdot \frac{1+\ln x}{2}=2\cdot 0$

Simplify fraction in left side of the equation

  $1+\ln x=0$

Subtract 1 on both sides of the equation

  $1+\ln x-1=0-1$

Simplifying on both sides

  $\ln x=-1$

Rewrite the logarithmic equation as an exponential equation

  $x=e^{-1}$

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

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

Graph:

  

Interpretation:

Setting up left side of the equation $f\left(x\right)=\frac{1+\ln x}{2}$ 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 $\frac{1+\ln x}{2}=0$

Conclusion:

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