solve for x
log(x-1)=log(x+2)-log(x+2)
$\displaystyle \log{(x-1)}=\log{(x+2)}-\log{(x+2)}$
$\displaystyle \log{(x-1)}=\log{\frac{(x+2)}{(x+2)}}$
$\displaystyle \log{(x-1)}=\log{1}$
$\displaystyle x-1=1$
$\displaystyle x=2$
$\displaystyle \log{(x-1)}=\log{(x+2)}-\log{(x+2)}$
$\displaystyle \log{(x-1)}=0$
$\displaystyle x-1=1$
$\displaystyle x=2$
I also feel bizarre.