solve for x
log(x-1)=log(x+2)-log(x+2)

2. Originally Posted by spensersievers
solve for x
log(x-1)=log(x+2)-log(x+2)
$\log{(x-1)}=\log{(x+2)}-\log{(x+2)}$
$\log{(x-1)}=\log{\frac{(x+2)}{(x+2)}}$
$\log{(x-1)}=\log{1}$
$x-1=1$
$x=2$

$\log{(x-1)}=\log{(x+2)}-\log{(x+2)}$
$\log{(x-1)}=0$
$x-1=1$
$x=2$

I also feel bizarre.

3. Originally Posted by spensersievers
solve for x
log(x-1)=log(x+2)-log(x+2)
The right hand side is rather bizarre, are you sure that is copied correctly?

-Dan