The forum discussion focuses on solving the system of equations defined by $\sqrt{x^2-2x+6}\cdot \log_{3}(6-y) = x$, $\sqrt{y^2-2y+6}\cdot \log_{3}(6-z) = y$, and $\sqrt{z^2-2z+6}\cdot \log_{3}(6-x) = z$. Participants agree that due to the symmetry of the equations, assuming $x = y = z$ simplifies the problem, leading to the solution $x = y = z = 3$. However, it is noted that while this is a valid solution, symmetry does not imply it is the only solution.
PREREQUISITESMathematicians, students studying algebra, and anyone interested in solving complex systems of equations involving logarithmic and square root functions.
Due to the symmetry, I assume that x = y = z.\text{Solve the following system of equations in real numbers:}
. . \begin{array}{ccc}\sqrt{x^2-2x+6}\cdot \log_{3}(6-y) &=&x \\<br /> \sqrt{y^2-2y+6}\cdot \log_{3}(6-z) &=& y \\<br /> \sqrt{z^2-2z+6}\cdot\log_{3}(6-x)&=&z\end{array}
Symmetry only guarantees that any permutation of the values of x, y, z for a solution is also a solution.soroban said:Hello, jacks!
I agree with pickslides . . .
Due to the symmetry, I assume that x = y = z.
Then we have: .\sqrt{x^2-2x+6}\cdot \log_3(6-x) \:=\:x
By inspection, we see that: .x\,=\,3.