MHB Solve System of Equations with Real Numbers

[juantheron]

Solve the following system of equations in real numbers:$\sqrt{x^2-2x+6}\cdot \log_{3}(6-y) =x$$\sqrt{y^2-2y+6}\cdot \log_{3}(6-z)=y$$\sqrt{z^2-2z+6}\cdot\log_{3}(6-x)=z .$

 

[pickslides]

Have you looked at (3,3,3) being a solution, not sure myself but it might work given the base of those logs.

 

[soroban]

Hello, jacks!

I agree with pickslides . . .

\text{Solve the following system of equations in real numbers:}

. . \begin{array}{ccc}\sqrt{x^2-2x+6}\cdot \log_{3}(6-y) &amp;=&amp;x \\<br /> \sqrt{y^2-2y+6}\cdot \log_{3}(6-z) &amp;=&amp; y \\<br /> \sqrt{z^2-2z+6}\cdot\log_{3}(6-x)&amp;=&amp;z\end{array}

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.

 

[CaptainBlack]

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.

Symmetry only guarantees that any permutation of the values of x, y, z for a solution is also a solution.

Obviously x=y=z=3 is a solution, but symmetry alone does not force us to conclude that it is the only solution.

CB