Question

If \(x^{a} y^{b}=e^{m}, x^{c} y^{d}=e^{n}\) \(\mathrm{D}_{1}=\left|\begin{array}{ll}\mathrm{m} & \mathrm{b} \\ \mathrm{n} & \mathrm{d}\end{array}\right|, \mathrm{D}_{2}=\left|\begin{array}{ll}\mathrm{a} & \mathrm{m} \\ \mathrm{c} & \mathrm{n}\end{array}\right|\) and \(\mathrm{D}=\left|\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right|\) then \(\mathrm{x}=\quad\) and \(\mathrm{y}=\) (a) \(\log \left(\overline{ \left.D_{1} / D\right)}, \log \left(\mathrm{D}_{2} \overline{/ D)}\right.\right.\) (b) \(\left(\mathrm{D}_{1} / \mathrm{D}\right),\left(\mathrm{D}_{2} / \mathrm{D}\right)\) (c) \(e^{[\\{(D) 1\\} / D]}, e^{[\\{(D) 2\\} / D]}\) (d) \(\left(\mathrm{D}_{2} / \mathrm{D}_{1}\right),\left(\mathrm{D}_{1} / \mathrm{D}_{1}\right)\)

Short answer

\(\mathrm{x} = e^{[\frac{D1}{D}]}, \mathrm{y} = e^{[\frac{D2}{D}]}\)

Step-by-step solution

Option (a)

\(\log \left(\overline{ \left.D_{1} / D\right)}, \log \left(\mathrm{D}_{2} \overline{/ D)}\right.\right.\) This option involves taking logarithms of the ratios \(D_1/D\) and \(D_2/D\). However, the given equations are exponential, and a logarithmic function doesn't seem to suit the situation.

Option (b)

\(\left(\mathrm{D}_{1} / \mathrm{D}\right),\left(\mathrm{D}_{2} / \mathrm{D}\right)\) This option involves using the ratios of the determinants directly. Since we have two-linear equations in the form of \(ax+by=m\) and \(cx+dy=n\), the determinants could be related to Cramer's rule for solving linear equations, which involves finding the ratios of determinants to obtain the solutions for x and y. This seems like a viable option.

Option (c)

\(e^{[\\{(D) 1\\} / D]}, e^{[\\{(D) 2\\} / D]}\) This option involves calculating the exponential functions based on the ratios of the (D) terms. Given that the original equations are exponential in nature, this could also be a possible solution for the problem.

Option (d)

\(\left(\mathrm{D}_{2} / \mathrm{D}_{1}\right),\left(\mathrm{D}_{1} / \mathrm{D}_{1}\right)\) This option involves finding the solution by just the ratio of \(D_2\) and \(D_1\) for x and the ratio of \(D_1\) to itself for y, which is always equal to 1. As the problem deals with higher powers and exponential equations, it doesn't seem to be an appropriate approach. Based on the analysis, option (b) and (c) seem to be the most promising. Since the problem is about exponential equations, we recommend that option (c) is most likely the correct choice: \(\mathrm{x} = e^{[\frac{D1}{D}]}, \mathrm{y} = e^{[\frac{D2}{D}]}\)

Key concepts

Exponential Equations

Exponential equations are mathematical expressions where the variable appears in the exponent. Understanding these equations is crucial in various domains of math and science, where growth processes are studied, such as in compounding interest, population growth, and radioactive decay. To solve an exponential equation, one often needs to manipulate the equation to have the same base on both sides, or alternatively, to use logarithms to bring down the exponent.

When dealing with systems of exponential equations like in the original exercise, the solutions typically involve a combination of algebraic methods and properties of exponents. In some instances, it’s possible to use transformations that reduce the exponential terms to a more straightforward system that can be approached with linear algebra techniques.

System of Equations

A system of equations is a set of two or more equations with the same variables. The goal is to find the values of these variables that satisfy all equations in the system simultaneously. There are several methods to solve such systems, including graphing, substitution, elimination, and using matrices.

In the context of our problem, we need to find the values of the variables that satisfy two exponential equations. While not as straightforward as linear systems, certain algebraic manipulations can render them solvable using similar principles, particularly when those manipulations lead us to a form where Cramer's Rule applies.

Determinants

Determinants are a special kind of function that take a square matrix and convert it into a single number. This number tells us a lot about the properties of the matrix and the linear system it corresponds to. For example, a non-zero determinant indicates that the system has a unique solution.

In the case of the exercise, determinants are used to assess the ratios of the exponents in the system of equations. Using the determinant values obtained from Cramer’s Rule, one can possibly solve for the variables in a system of linear equations. Despite the exponential nature of the equations at hand, transformations and use of determinants allow for a possible solution path.

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear mappings between these spaces. It includes the study of lines, planes, and subspaces, but is also concerned with properties common to all vector spaces. A key tool in linear algebra is matrices, which can be used to solve systems of linear equations.

Cramer's Rule, which is a theorem in linear algebra, is instrumental in solving systems of linear equations via the use of determinants. This rule gives us a concrete method to find a unique solution to a system, provided that the determinant of the system's matrix is not zero. By extending these concepts to handle an exponential system, as we are trying to do in the exercise presented, we apply the clear-cut methods of linear algebra to the more complex setting of exponential equations.