Chapter 3.4, Problem 64E

To determine

To find : solution to the equationgraphically and algebraically.

Answer to Problem 64E

The solution to the equation $6e^{1-x}=25$ is $\underset{\_}{\boxed{x=3.328}}$

Explanation of Solution

Given information: $6e^{1-x}=25$

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 $f\left(x\right)=6e^{1-x}$ and right side of the equation $g\left(x\right)=25$ .

The xcoordinate of point of intersection of two graphs is the solution to the equation $6e^{1-x}=25$

Formula Used:

1. $\ln m^n=n\ln m$
2. $\ln e=1$

Calculation:

Divide by 6 on both sides of the equation

  $\frac{6e^{1-x}}{6}=\frac{25}{6}$

Simplify fraction on both sides of the equation

  $e^{1-x}=\frac{25}{6}$

Take natural logarithm on both sides of the equation

  $\ln e^{1-x}=\ln \left(\frac{25}{6}\right)$

Applying the logarithmic property $\ln m^n=n\ln m$ to rewrite the left side of the equation

  $\left(1-x\right)\cdot \ln e=\ln \left(\frac{25}{6}\right)$

Replace 1 for $\ln e$

  $1-x=\ln \left(\frac{25}{6}\right)$

Subtract 1 on both sides

  $1-x-1=\ln \left(\frac{25}{6}\right)-1$

Simplify equation on both sides of the equation

  $-x=\ln \left(\frac{25}{6}\right)-1$

Multiply -1 throughout the equation

  $-1\left(-x\right)=-1\left(\ln \left(\frac{25}{6}\right)-1\right)$

Distribute negative in right side of the equation

  $x=-\ln \left(\frac{25}{6}\right)+1$

Simplify the right side of the equation

  $x\approx -0.4271164$

Conclusion:

The solution to the equation $6e^{1-x}=25$ is $\underset{\_}{\boxed{x=-0.427}}$