Question 63486
<b>log3 (x) + log 3 (x-2) = log 3 (8)</b>

log3(x)+log3(x-2) = log3(x(x-2)) = log3(8).

So, x(x-2) = 8, or x^2-2x-8=0.

This factors into (x+2)(x-4) = 0.
So, x=-2 or x=4. 
But, if x=-2 then you're trying to get the log of a negative number. That's not defined.

So, {{{highlight(x=4)}}} is the solution.