Question

Solve the system: (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, and z.)﻿{2ln⁡w+ln⁡x+3ln⁡y−2ln⁡z=−64ln⁡w+3ln⁡x+ln⁡y−ln⁡z=−2ln⁡w+ln⁡x+ln⁡y+ln⁡z=−5ln⁡w+ln⁡x−ln⁡y−ln⁡z=5\begin{cases}
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{cases}⎩⎨⎧​2lnw+lnx+3lny−2lnz=−64lnw+3lnx+lny−lnz=−2lnw+lnx+lny+lnz=−5lnw+lnx−lny−lnz=5​

Accepted Answer

Step 1: Introduce the substitutions as given: let $$A = \ln w$$, $$B = \ln x$$, $$C = \ln y$$, and $$D = \ln z. $$Rewrite the system of equations in terms of $$A$$, $$B$$, $$C$$, and $$D. $$Step 2: Rewrite each equation by replacing $$\ln w$$ with $$A$$, $$\ln x$$ with $$B$$, $$\ln y$$ with $$C$$, and $$\ln z$$ with $$D. $$The system becomes:

$$\begin{cases}
2A + B + 3C - 2D = -6 \
4A + 3B + C - D = -2 \
A + B + C + D = -5 \
A + B - C - D = 5
\end{cases}$$ Step 3: Solve this system of linear equations for $$A$$, $$B$$, $$C$$, and $$D$$ using methods such as substitution, elimination, or matrix operations (e.g., Gaussian elimination). Step 4: Once you find the values of $$A$$, $$B$$, $$C$$, and $$D$$, recall that these are logarithms of the original variables. Use the inverse logarithm (exponentiation) to find $$w$$, $$x$$, $$y$$, and $$z$$:

$$w = e$$^{A}$$, x$$ = $$e^{B}$$, $$y = e$$^{C}$$, z$$ = $$e^{D}. $$Step 5: Verify your solutions by substituting $$w$$, $$x$$, $$y$$, and $$z$$ back into the original system of equations to ensure all equations are satisfied.

Verified video answer for a similar problem:

Key Concepts

Logarithmic Properties and Transformations

Solving Systems of Linear Equations

Exponentiation to Reverse Logarithms