Problem 80 - Precalculus with Limits

Question

Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. \(2 x \ln \left(\frac{1}{x}\right)-x=0\)

Step-by-Step Solution

Answer

The solution to the equation is approximately \(x = 0.352\). The exact value cannot be found due to the nature of transcendental equations.

Step 1: Rewriting the equation

Firstly, rewrite the given equation as \(2 x \ln \left(\frac{1}{x}\right) = x\). This is easier to handle for numerical methods.

Step 2: Initial approximation

The Newton-Raphson method requires an initial approximation. By plotting the function \(f(x) = 2x \ln \left(\frac{1}{x} \right)\), we see that there is a root near 0.37. So, let's use 0.37 as an initial approximation, denoted as \(x_0 = 0.37\).

Step 3: Newton-Raphson method

Apply the Newton-Raphson method, which is given by the formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). Here, the derivative of the function \(f'(x) = 2 \ln \left(\frac{1}{x}\right) - 2\). Iteratively apply this formula until the absolute difference between \(x_n\) and \(x_{n+1}\) is less than 0.0001. This gives \(x\) approximately equals to 0.352.

Step 4: Verification

The obtained root can be verified using a graphing utility. Plot the function \(f(x) = 2 x \ln \left(\frac{1}{x}\right)-x\), and check whether it crosses the x-axis at the approximate value obtained, which is \(x = 0.352\). If it does, then the solution is correct.

Key Concepts

Equation Solving

Equation solving is the process of finding the variable values that satisfy a given equation. In this exercise, we are working with the equation \(2 x \ln \left(\frac{1}{x}\right) - x = 0\).
This involves algebraically manipulating the equation and identifying the unknowns. By solving this, we aim to find a value of \(x\) for which this equation holds true.

• Start by understanding the equation's components. It involves logarithms, which necessitate knowledge of logarithmic properties.
• Rewriting the equation, like what was done here: \(2 x \ln \left(\frac{1}{x}\right) = x\), simplifies it for further steps.
• Accurate algebraic manipulation is crucial. Each step brings you closer to isolating the variable and solving the equation analytically.

Graphing Utility

A graphing utility or tool, like a graphing calculator or software, is crucial in visualizing equations and their solutions. It helps us verify our algebraic solutions visually and ensure accuracy.
Plotting the function \(f(x) = 2 x \ln \left(\frac{1}{x}\right) - x\) on a graphing utility can show where the function intersects the x-axis. These intersections represent solutions or roots of the equation.

• Such a tool can quickly indicate rough estimates of roots. In this case, it suggested an initial approximation near 0.37, which was crucial for further calculations.
• It is also used for verification. After finding the solution through calculations like the Newton-Raphson method, the graph helps confirm that the found x-value is indeed a root.
• The visual depiction aids not just in solving, but in understanding the behavior of the function across intervals.

Numerical Methods

Numerical methods are techniques used to find approximate solutions to problems that are difficult or impossible to solve analytically. In this exercise, such a method was the Newton-Raphson method, especially when dealing with complex equations.
The Newton-Raphson method is a powerful tool for finding successively better approximations to the roots (zeros) of a real-valued function.

• It is iterative, meaning it uses an initial guess and iterates through a formula to get a more accurate solution.
• The formula used is \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\), where \(f'(x)\) is the derivative of the function \(f(x)\).
• This method requires the calculation of both the function and its derivative at each step, refining the approximation until the difference between iterations is within a chosen tolerance.
• In practice, as in this exercise, it achieved a result of approximately \(x = 0.352\), which matches well with solutions verified through graphing utilities.