Problem 98
Question
Use a graphing utility to solve the equation. Approximate the result to three decimal places. Verify your result algebraically.\(10-4 \ln (x-2)=0\)
Step-by-Step Solution
Verified Answer
The result of the equation will depend on the graphing utility you use, but the most important thing is to graph it correctly, approximate the value to three decimal places and verify it algebraically.
1Step 1: First step: Graphing the equation
To graph the equation \(10-4 \ln (x-2)=0\), a graphing utility, such as Desmos or GeoGebra, can be used. Enter the equation in the graphing utility and find the x-coordinate of the point where the graph intersects the x-axis. This will be the approximate solution.
2Step 2: Second step: Approximating the result
According to the requirement of the exercise, the result needs to be approximated to three decimal places. In the graphing utility, the coordinates are usually displayed with many decimal places. Therefore, round off the x-coordinate of the intersection point to three decimal places.
3Step 3: Third step: Algebraic verification
An algebraic verification of the result is required. This means, to substitute the approximated result back into the equation \(10-4 \ln (x-2)=0\) to verify the result. If it’s the root of the equation, the equation should yield 0 when the approximated root is substituted back in.
Key Concepts
Graphing UtilityApproximationAlgebraic Verification
Graphing Utility
Graphing utilities are powerful tools that help visualize equations and find solutions. For the equation \(10-4 \ln (x-2)=0\), a graphing utility assists by drawing the curve associated with the equation. The goal is to find where the curve crosses the x-axis.
This crossing point is crucial because it represents where the equation equals zero—the solution. Tools such as Desmos or GeoGebra are particularly popular for these tasks. They allow students to graph equations with ease and explore their properties. To utilize it here, you would input the equation into the utility, and observe the graph it generates.
Look for the point where the curve intersects the x-axis. This point's x-coordinate is your solution. The intersection reflects the solution to the equation and is key to understanding how graphing provides a visual means of solving equations.
This crossing point is crucial because it represents where the equation equals zero—the solution. Tools such as Desmos or GeoGebra are particularly popular for these tasks. They allow students to graph equations with ease and explore their properties. To utilize it here, you would input the equation into the utility, and observe the graph it generates.
Look for the point where the curve intersects the x-axis. This point's x-coordinate is your solution. The intersection reflects the solution to the equation and is key to understanding how graphing provides a visual means of solving equations.
Approximation
Once you've identified the x-coordinate where the graph intersects the x-axis, you have a preliminary solution. However, to meet the exercise requirements, this solution needs to be presented in a specific numerical form—three decimal places.
Graphing utilities typically display coordinates with high precision which includes many decimal places. To approximate, simply round the x-coordinate to three decimal places. This step ensures the solution is expressed in a clear, concise, and standardized format.
For example, if the graphing utility provides an x-coordinate of 4.123456, you would round this to 4.123. This approximation is vital because it aligns with the precision required in many mathematical contexts and communicates the solution effectively.
Graphing utilities typically display coordinates with high precision which includes many decimal places. To approximate, simply round the x-coordinate to three decimal places. This step ensures the solution is expressed in a clear, concise, and standardized format.
For example, if the graphing utility provides an x-coordinate of 4.123456, you would round this to 4.123. This approximation is vital because it aligns with the precision required in many mathematical contexts and communicates the solution effectively.
Algebraic Verification
Solving an equation visually with a graphing utility is insightful but verifying the accuracy of that solution through algebra is equally important. Here’s why algebraic verification is essential. It not only checks the correctness of your approximate solution but also deepens your understanding of the equation.
To verify the approximation algebraically, substitute your found solution, say \(x=4.123\), back into the original equation \(10-4 \ln (x-2)=0\). Use properties of logarithms and basic algebra to simplify the equation and check whether it equals zero.
If the solution satisfies the equation, congratulations, you've verified your result! Algebraic verification further bolsters confidence in your solution and affirms the practical use of graphing utilities as tools for solving complex equations. This step ensures the result is not only visually but also factually accurate.
To verify the approximation algebraically, substitute your found solution, say \(x=4.123\), back into the original equation \(10-4 \ln (x-2)=0\). Use properties of logarithms and basic algebra to simplify the equation and check whether it equals zero.
If the solution satisfies the equation, congratulations, you've verified your result! Algebraic verification further bolsters confidence in your solution and affirms the practical use of graphing utilities as tools for solving complex equations. This step ensures the result is not only visually but also factually accurate.
Other exercises in this chapter
Problem 97
Use a graphing utility to solve the equation. Approximate the result to three decimal places. Verify your result algebraically.\(3-\ln x=0\)
View solution Problem 97
Condense the expression to the logarithm of a single quantity.\(\log _{8}(x-2)-\log _{8}(x+2)\)
View solution Problem 98
Condense the expression to the logarithm of a single quantity.\(3 \log _{7} x+2 \log _{7} y-4 \log _{7} z\)
View solution Problem 99
Find the time required for a \(\$ 1000\) investment to double at interest rate \(r\), compounded continuously.\(r=0.0625\)
View solution