$\displaystyle log(x-2)+log2=2logy;\ log(x-3y+3)=0$ find out the value of (x) and (y).
Note that $\displaystyle \log(x-2) + \log 2 = \log(2x-4)$, and that $\displaystyle 2 \log y = \log y^2$. Therefore the first equation becomes
$\displaystyle \log(2x-4) = \log y^2 \Rightarrow 2x-4 = y^2$
The second equation, note that $\displaystyle \log 1 = 0$ (for any non-zero base), so $\displaystyle x - 3y + 3 = 0$.
Solve the system of equations.