logarithms please help

spensersievers · March 18th 2008, 06:01 PM

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

**SengNee** · March 18th 2008, 06:10 PM

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.

**topsquark** · March 18th 2008, 06:24 PM

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

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