Chapter 3.4, Problem 68E

To determine

To find : solution to the equationgraphically and algebraically.

Answer to Problem 68E

The solution to the equation $10-4\ln \left(x-2\right)=0$ is $\underset{\_}{\boxed{x=14.182}}$

Explanation of Solution

Given information: $10-4\ln \left(x-2\right)=0$

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.

Graph:

  

Interpretation:

Setting up left side of the equation $f\left(x\right)=10-4\ln \left(x-2\right)$ 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 $10-4\ln \left(x-2\right)=0$

Formula Used:

1. If logarithmic equation is $\ln A=m$ , then the equivalent exponential equation is $A=e^m$
2. $\ln e=1$

Calculation:

Subtract 10 on both sides of the equation

  $10-4\ln \left(x-2\right)-10=0-10$

Simplify fraction on both sides

  $-4\ln \left(x-2\right)=-10$

Divide -4 on both sides of the equation

  $\frac{-4\ln \left(x-2\right)}{-4}=\frac{-10}{-4}$

Simplifying fraction on both sides

  $\ln \left(x-2\right)=2.5$

Rewrite the logarithmic equation to exponential equation

  $x-2=e^{2.5}$

Adding 2 on both sides of the equation

  $x-2+2=e^{2.5}+2$

Simplifyin right side of the equation

  $x\approx 14.182$

Conclusion:

The solution to the equation $10-4\ln \left(x-2\right)=0$ is $\underset{\_}{\boxed{x=14.182}}$