Problem 44 - College Algebra

Question

Solve the system: $$ \left\\{\begin{array}{cc} \ln w+\ln x+\ln y+\ln z= -1 \\ -\ln w+4 \ln x+\ln y-\ln z= 0 \\ \ln w-2 \ln x+\ln y-2 \ln z= 11 \\ -\ln w-2 \ln x+\ln y+2 \ln z =-3 \end{array}\right. $$

Step-by-Step Solution

Answer

The system of equations has no solution.

Step 1: Simplification using Logarithmic Rules

Rewrite each equation in a simpler form using the properties of logarithms. According to the property \(ln(A)+ln(B) = ln(AB)\), the given equations can be rewritten as: i) \(ln(wxyz) = -1\), ii) \(ln(x^4/wyz) = 0\), iii) \(ln(wy/(x^2 z^2)) = 11\), iv) \(ln(wy/(x^2 z^2)) = -3.\)

Step 2: Exponential Form Conversion

Since the logarithmic function and the exponential function are inverses of each other, convert the equations into exponential form: i) \(wxyz = e^{-1}\), ii) \(x^4/wyz = 1\), iii) \(wy/x^2 z^2 = e^{11}\), iv) \(wy/x^2 z^2 = e^{-3}.\)

Step 3: Simplification of Equations

From equation ii) it is evident that \(wyz = x^4\). Now, replace \(wyz\) by \(x^4\) in equations i), iii) and iv) to get: i) \(x^5 z = e^{-1}\), iii) \(x^2/(x^2 z^2) = e^{11}\), so, \(1/z^2 = e^{11}\), iv) \(x^2/(x^2 z^2) = e^{-3}\), so, \(1/z^2 = e^{-3}\). Since equations iii) and iv) are identical, the system is consistent only when \(e^{11} = e^{-3}\), which is not correct, therefore the system has no solution.

Key Concepts

Logarithmic Properties

Understading logarithmic properties is crucial when dealing with equations that involve logarithms. Logarithms, the inverse operations of exponentiation, have unique properties that make them easy to manipulate algebraically. The logarithm of a product is the sum of the logarithms of the individual factors, expressed as \(\ln(AB) = \ln(A) + \ln(B)\). Similarly, the logarithm of a quotient is the difference of the logarithms, \(\ln(A/B) = \ln(A) - \ln(B)\), and the logarithm of a power is the exponent times the logarithm of the base, \(\ln(A^k) = k\ln(A)\).

Using these properties can significantly simplify complex expressions and is a fundamental step in solving logarithmic equations. In the given exercise, the sum and difference of logarithms are consolidated into single logarithmic terms, reducing the system of equations to a more manageable form. Remember, handling logarithmic expressions with care, using the correct properties, can lead to accurate and concise solutions.

Exponential Form Conversion

Often, to solve logarithmic equations, we need to convert the logarithmic form to exponential form. This conversion leverages the fact that logarithms and exponentiation are inverse operations. For any logarithmic equation of the form \(\ln(a) = b\), the exponential form is \(e^b = a\). This step is integral when we are trying to isolate the variable and solve for it.

In the solved exercise, converting the logarithmic form to exponential form allows us to restate the equation in a more approachable way, turning a complex logarithmic relationship into a simple exponential equation. This is a powerful step in the problem-solving process, as it typically leads to equations that are more straightforward to solve. Understanding how to fluidly switch between logarithmic and exponential forms is essential for anyone looking to master the topic of logarithms.

Logarithmic Equations

Solving logarithmic equations often involves using logarithmic properties initially and then converting to exponential form as an intermediate step. Logarithmic equations contain variables inside the logarithm. A systematic approach to solving these involves simplifying using logarithmic properties, isolating the logarithmic part, converting to exponential form, and then finding the values of the unknown variables.

However, it's important to note that not all systems of logarithmic equations will have a solution, as is the case in the given problem. When two equations result in exponential functions with different bases that are not equal (e.g., \(e^{11}\) versus \(e^{-3}\)), there's a contradiction, indicating that the system has no solution. Recognizing such inconsistencies is a critical part of problem-solving with logarithmic equations.