Hi Jensen;
log_{2}(2x+6)log_{2}(x1)=log_{2}3
Our first priority is to verify that all logs are to the same base. These are all to the base of 2. We may now proceed.
The logarithmprinciple we need to apply is...
log a/b=log a  log b
Therefore...
log_{2}[(2x+6)/(x1)]=log_{2}3
We can eliminate log_{2} on both sides...
(2x+6)/(x1)=3
(2x+6)/(x1)=3/1
Let's crossmultiply...
1(2x+6)=3(x1)
(2x+6)=3(x1)
2x+6=3x3
Let's combine like terms...
Let's subtract 2x from both sides...
2x+2x+6=3x32x
6=x3
Let's add 3 to both sides...
3+6=x3+3
9=x
Let's check our work by returning to the original equation...
log_{2}(2x+6)log_{2}(x1)=log_{2}3
log_{2}[(2*9)+6)log_{2}(91)=log_{2}3
log_{2}(18+6)log_{2}(8)=log_{2}3
log_{2}(24)log_{2}(8)=log_{2}3
log a/b=log a  log b
log_{2}[(24)/(8)]=log_{2}3
log_{2}3=log_{2}3
2/13/2014

Vivian L.