Problem 43 - Algebra and Trigonometry

Question

Solve the system: $$ \left\\{\begin{array}{c} {2 \ln w+\ln x+3 \ln y-2 \ln z=-6} \\ {4 \ln w+3 \ln x+\ln y-\ln z=-2} \\ {\ln w+\ln x+\ln y+\ln z=-5} \\ {\ln w+\ln x-\ln y-\ln z=5} \end{array}\right. $$ (Hint: Let \(A=\ln w, B=\ln x, C=\ln y,\) and \(D=\ln z .\) Solve the system for \(A, B, C,\) and \(D .\) Then use the logarithmic equations to find \(w, x, y, \text { and } z .)\)

Step-by-Step Solution

Answer

The solutions are \( w = e, x = 1/e^2, y = 1/e, z = 1/e^3 \)

Step 1: Substitution

Given that \( A = \ln{w}, B = \ln{x}, C = \ln{y}, D = \ln{z} \), substitute these into the original equations resulting in the following system of linear equations: \[\begin{align}2A + B + 3C - 2D &= -6,\ 4A + 3B + C - D &= -2,\A + B + C + D &= -5,\ A + B - C - D &= 5.\end{align}\]

Step 2: Solution of the System of Linear Equations

These can be solved using a method such as Gaussian elimination or substitution method. Solving them get the following values for \( A, B, C, D \): \[\begin{align}A = 1,\ B = -2,\C = -1,\ D = -3.\end{align}\]

Step 3: Convert Back to Original Variables

Recall from the definition of the natural logarithm that \( A = \ln{w}, B = \ln{x}, C = \ln{y}, D = \ln{z} \). This implies that \( w = e^A, x = e^B, y = e^C, z = e^D \). Substituting the values calculated in the previous step leads to: \[\begin{align}w = e^1 = e,\ x = e^{-2} = 1/e^2,\y = e^{-1} = 1/e,\ z = e^{-3} = 1/e^3.\end{align}\]

Key Concepts

Natural Logarithm

The natural logarithm, denoted by \( \ln \), is a logarithm to the base \( e \), where \( e \) is an irrational number approximately equal to 2.71828. It's a fundamental concept in algebra and calculus due to its properties and its relationship with growth processes.

In the context of solving equations, the natural logarithm is used to transform exponential equations into linear ones, making them easier to work with. The inverse property \( \ln(e^x) = x \) simplifies the conversion between exponential and linear forms, a crucial aspect in solving systems of equations involving logarithms.

• \( \ln(a) + \ln(b) = \ln(ab) \): This property helps combine products into a single logarithm, simplifying many algebraic processes.
• \( \ln(a) - \ln(b) = \ln(a/b) \): Useful for converting divisions inside a logarithm into subtraction, aiding in simplifying expressions.

Understanding these properties is essential when dealing with equations that involve logarithms, like in the problem where we replace \( w, x, y, \text{and } z \) with \( \ln w, \ln x, \ln y, \text{and } \ln z \), respectively, to linearize exponential relationships.

Gaussian Elimination

Gaussian elimination is a powerful method used for solving systems of linear equations. It involves three types of row operations to transform a matrix into its row-echelon form and then into reduced row-echelon form.

Here's a summary of the process involved in Gaussian elimination:

• Forward Elimination: This step transforms the system of equations into an upper triangular matrix. It involves making the elements below the pivot positions zero using row operations.
• Back Substitution: Once the matrix is in upper triangular form, solve each equation starting from the last one and substitute upwards to find the other variables.

In the given exercise, Gaussian elimination can be employed to simplify and solve the system of equations obtained from substituting \( A, B, C, \) and \( D \). This method is particularly useful for larger systems or when precise solutions are needed.

Substitution Method

The substitution method is another effective approach for solving systems of linear equations. It is often favored for its simplicity, especially when dealing with smaller systems or when variables can be easily isolated.

Here's how it generally works:

• Choose one equation and solve for one variable in terms of the others.
• Substitute this expression into the other equations, replacing the selected variable.
• Continue the process until you solve for all variables.

In the context of the exercise, once transformed into a system of linear equations using \( A, B, C, \) and \( D \), you can opt to solve for one variable at a time, making the computations straightforward. This method aligns neatly with substituting back the values to the logarithmic expressions to find \( w, x, y, \text{and } z \).

Linear Equations

Linear equations are fundamental mathematical expressions that describe a straight line when graphed. They take the general form \( ax + by + cz + \ldots = d \), where \( a, b, \) and \( c \) are coefficients, and \( d \) is the constant term. These equations are essential in various fields, including algebra, calculus, and engineering.

In the context of a system of equations, each equation represents a geometric plane, and the solution to the system is at the intersection of these planes. For example, in a two-dimensional space, this would be the intersection of lines, while in three dimensions, it would be the intersection of planes.

• Solving a system of linear equations helps find values that satisfy all equations simultaneously, critical in applications like optimization, model predictions, and data fitting.
• Techniques such as substitution and Gaussian elimination can convert complex problems into simple linear systems, making them easier to tackle.

The exercise transforms logarithmic equations into a linear system in variables \( A, B, C, \) and \( D \), allowing for the application of linear solving techniques to eventually deduce values for \( w, x, y, \text{and } z \).