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System of equations

jacks

Well-known member
Apr 5, 2012
226
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

Member
Feb 1, 2012
57
Have you looked at (3,3,3) being a solution, not sure myself but it might work given the base of those logs.
 

soroban

Well-known member
Feb 2, 2012
409
Hello, jacks!

I agree with pickslides . . .


[tex]\text{Solve the following system of equations in real numbers:}[/tex]

. . [tex]\begin{array}{ccc}\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\end{array}[/tex]

Due to the symmetry, I assume that [tex]x = y = z.[/tex]

Then we have: .[tex]\sqrt{x^2-2x+6}\cdot \log_3(6-x) \:=\:x[/tex]

By inspection, we see that: .[tex]x\,=\,3.[/tex]
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Hello, jacks!

I agree with pickslides . . .



Due to the symmetry, I assume that [tex]x = y = z.[/tex]

Then we have: .[tex]\sqrt{x^2-2x+6}\cdot \log_3(6-x) \:=\:x[/tex]

By inspection, we see that: .[tex]x\,=\,3.[/tex]

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
 
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