Solve simultaneous equations $\log(x-2)+ \log 2=2 \log y$, $\log (x-3y+3)=0$ (Not sure of solutions in book)

Question

Solve simultaneous equations $\log(x-2)+\log2=2\log y$, $\log(x-3y+3)=0$ (Not sure of solutions in book)

My method: $\log(x-2)+\log 2-\log y^2=0 \Rightarrow \log\left(\frac{x}{y^2}\right)=0 \Rightarrow y^2=x$

$\log(x-3y+3)=0 \Rightarrow y^2-3y+2=0 \Rightarrow (y-1)(y-2)=0$

When $y=1,x=1; y=2,x=4$

Book solutions: When $y=4,x=10;y=2,x=4$

Answer 1

HINT: the first equation is $\log(2(x-2))=\log(y^2)$ thus we have $$2(x-2)=y^2$$ the second equation is equivalent to $$x-3y+3=1$$ Can you proceed? from the second equation we get $$y=\frac{x+2}{3}$$ plugging this in the first one we obtain $$2x-4=\left(\frac{x+2}{3}\right)^2$$

Answer 2

You might want to read this section first, https://en.wikipedia.org/wiki/List_of_logarithmic_identities#Using_simpler_operations.

$1) \log(x-2)+\log2=2\log y $

$2) \log(x-3y+3)=0 $

$1) \log(x-2)+\log2=\log(2x-4)=\log y^2 \iff 2x-4 = y^2$

$2) \log(x-3y+3)=0 \iff x-3y+3=1$

by taking exponentials.

$1) x=\frac{y^2}{2}+2$

$2) x-3y+3=1$

subbing in $\frac{y^2}{2}+2-3y+3=1 \iff \frac{y^2}{2}-3y+4=0 \iff y^2-6y+8=0$

the rest will be easy now...