MHB Integer ordered pairs in logarithmic equation

AI Thread Summary
The discussion centers on finding integer ordered pairs (x, y, z) that satisfy a set of logarithmic equations. The constraints indicate that x, y, and z must be less than 6, leading to the transformation of the logarithmic terms into fractions involving square roots. A participant suggests that the only integer solution may be the trivial case where x, y, and z all equal 3, based on the nature of logarithmic functions. The conversation highlights the complexity of the equations and the challenge in determining additional solutions. Ultimately, the focus remains on the potential uniqueness of the solution at (3, 3, 3).
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no. of integer ordered pairs of $(x,y,z)$ in

$ \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 $

My approach :: Here $6-x,6-y,6-z>0$. So $x,y,z<6$

Now $\displaystyle \log_{3}(6-y) = \frac{x}{\sqrt{x^2-2x+6}}=\frac{x}{\sqrt{(x-1)^2+5}}$

and $\displaystyle \log_{3}(6-z) = \frac{y}{\sqrt{y^2-2y+6}}=\frac{y}{\sqrt{(y-1)^2+5}}$

and $\displaystyle \log_{3}(6-x) = \frac{z}{\sqrt{z^2-2z+6}}=\frac{z}{\sqrt{(z-1)^2+5}}$

How can I calculate after that.

Help please

Thanks
 
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Re: integer ordered pairs in logarithmic equation

I don't have a reference for this, but I guess that if $n$ is an integer then $\log_3n$ is transcendental unless $n$ is a power of $3$. If so, then the only solution to those equations must be the obvious one $x=y=z=3$.
 
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