Problem 53 - Precalculus with Limits

Question

Choosing a Solution Method In Exercises \(49-56\) , solve the system graphically or algebraically. Explain your choice of method. $$\left\\{\begin{array}{l}{y-e^{-x}=1} \\ {y-\ln x=3}\end{array}\right.$$

Step-by-Step Solution

Answer

The problem involves solving a system of equations that include exponential and logarithmic functions. The solution process involves both graphical and algebraic methods. The graphical method helps to visualize the problem and guess the solution, while the algebraic method offers precision. The exact solution depends on the results of the iterations and might need a computer program or calculator to solve.

Step 1: Graphical Representation

We can start by sketching the graph of each function. The graph of \(y-e^{-x}=1\) can be rewritten and plotted as \(y=e^{-x}+1\). The graph of \(y-\ln x=3\) can be also rewritten and plotted as \(y=\ln x +3\). We can observe that these two functions intersect at a certain point, which represents the solution to the system of equations.

Step 2: Solving Algebraically

While graphical solutions can give an idea of the possible solutions, they might not be precise. For an accurate answer, we can solve the system algebraically. Equate the two equations to find \(x\). That gives us \(e^{-x}+1=\ln x +3\). To find the exact value of \(x\), one might use iteration or numerical methods, such as the Newton-Raphson method, since this equation is not simple to solve.

Step 3: Determining y

Once we find the value of \(x\), we can substitute it into either one of the original equations to find the corresponding value of \(y\). For example, if we use the first equation, we substitute \(x\) into \(y=e^{-x}+1\), and calculate the value of \(y\).

Key Concepts

Graphical Representation

Understanding the graphical representation of a system of equations is a fundamental aspect of algebra that provides visual insights into the behavior of equations. It involves plotting each equation on a coordinate plane and observing where they intersect. The point of intersection represents the solution to the system because it is the set of coordinates that satisfy all equations involved.

For the given system, we first express the equations in terms of y: the first becomes \(y=e^{-x}+1\) and the second \(y=\ln x +3\). Plotting these on a graph reveals their behavior: the exponential decay of \(e^{-x}\) shifting upward by 1 and the natural logarithm graph shifting upward by 3. The graphical solution is approximate but provides an immediate visual answer, and it's particularly useful when the algebraic solution is complex. It is ideal for understanding the general region where the solution lies and for verifying algebraic or numeric solutions.

However, graphing might not always yield the precision needed for exact solutions or when dealing with complex equations. Furthermore, when using the graphing method, it's important to take into account the scale and the resolution of the graph so that the point of intersection can be correctly identified.

Algebraic Solution

Conversely, an algebraic solution entails manipulating the equations using algebra to find an exact solution. This solution provides a precise answer and works well for both linear and non-linear systems. It is achieved by applying algebraic techniques such as substitution, elimination, or using advanced methods for non-linear equations.

For the exercise at hand, after equating \(e^{-x}+1\) to \(\ln x +3\), we are prompted to undertake the challenging task of isolating \(x\) algebraically. This is where traditional methods fall short, and iterative numerical methods become useful. An exact solution through algebraic manipulation here is impractical as it involves transcendental functions that do not have closed-form solutions.

Applying algebraic solutions can provide a strong foundation for understanding the properties and relationships within a system of equations. It's crucial for students to learn these methods as they offer the skills necessary to tackle a wide range of problems, including solving equations for unknown variables, simplifying expressions, and more. The beauty of algebra lies in its ability to provide clear, definitive answers and its wide applicability in various fields of study.

Newton-Raphson Method

When facing equations that are difficult or impossible to solve algebraically, the Newton-Raphson method comes to the rescue as a powerful numerical technique. It is an iterative method used to find successively better approximations to the roots (or zeros) of a real-valued function.

The basic idea is to start with an initial guess for the root and then refine that guess by applying the Newton-Raphson formula:\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\] Where \(f(x)\) is the function for which we are trying to find the root, \(f'(x)\) is its derivative, and \(x_n\) is the current guess. The process is repeated until the change in \(x\) becomes negligible, indicating that we have arrived at a solution that is accurate enough for our needs.

For our specific equation \(e^{-x}+1=\ln x +3\), the task is to define \(f(x)\) as this equation rearranged to one side, \(f(x) = e^{-x} - \ln x + 2\), and then to apply the Newton-Raphson iteration. This approach is particularly useful when dealing with complex relationships that cannot be neatly unfolded through algebra alone. It's also a staple in computational mathematics, where analytical solutions are not always available, and highlights the increasing importance of numerical methods in tackling real-world problems.