log(2x+6)=1+log(x-2)

Question

I am having trouble getting the right answer. I know you have to divide the  two-

log (2x-6)/(x-2)=1

I go about solving it, but then I don't get the right answer.

Answer 1

log(2x+6) -log(x-2)=1

=log[(2x+6) / (x-2)]=1

assuming log tu base=10, log(sumthun)=1 meen sumthun=10

so (2x+6) / (x-2)=10

2x-6=10*(x-2)

2x-6-10x+20=0

-8x=-14

x=14/8=1.75