Non-linear System of Equations

Let $x'=x, y'=y-1, z'=z-2$ . Then:

$x'+\log(x')=y'$

Substituting,

$y'+\log(y')=z'$

then again to get

$z'+\log(z')=x'$

and more substitution, this time substitute for the first $y'$ in the second equation then use that for first $z'$ in the last equation:

$x'+\log(x')+\log(y')+\log(z')=x'$

Now it's obvious that if $\log(x')=\log(y')=\log(z')=0$ then we have a solution hence $x'=1, y'=1, z'=1$ and $x=1, y=2, z=3$ . We know that $\log(x'), \log(y'), \log(z')\ge0$ : observe that if $\log(z')<0$ then $z'+\log(z')=x'$ hence $z'>x'$ and $\log(x')$ is also less than 0. This method can be used to produce a sort of 'chain reaction' meaning if any one of them is negative we can use this repeatedly to show that all are negative. So the solution found is unique.

It is given that:

$\begin{cases} x + \log{(x)} & = y-1 &...(1) \\ y + \log{(y-1)} & = z-1 &...(2) \\ z + \log{(z-2)} & = x+2 &...(3) \end{cases}$

$(1)+(2)+(3):$

$\begin{aligned} x + y + z + \log{(x)} + \log{(y-1)} + \log{(z-2)} & = y+z+x -1-1+2 \\ \Rightarrow \log{(x)} + \log{(y-1)} + \log{(z-2)} & = 0 \\ \Rightarrow x(y-1)(z-2) & = 1 \end{aligned}$

We note that $x \ne -1$ , $y - 1 \ne -1$ and $z-2 \ne -1$ because $\log{(-1)}$ is undefined, there is only one trivial solution:

$\begin{cases} x = 1 & \Rightarrow x = 1 \\ y -1 = 1 & \Rightarrow y = 2 \\ z -2 = 1 & \Rightarrow z = 3 \end{cases}$

Since there is only one solution, therefore the trivial solution is the unique solution and $x+y+z = 1+2+3 = \boxed{6}$

Adding the three equations we get Log x + log (y-1)+ log(z-2) =0; Log(x)(y-1)(z-2)=0; X(y-1)(z-2)=1; Which has multiple solutions.. Where is this wrong? Please excuse the lack of formatting.