Assuming #log# is referring to the base-#10# logarithm, we can apply the properties that #log(a)+log(b) = log(ab)# and #10^log(x)=x# to obtain
#log(x)+log(x-3)=1#
#=> log(x(x-3)) = 1#
#=> 10^log(x(x-3)) = 10^1#
#=> x(x-3) = 10#
#=> x^2-3x - 10 = 0#
#=> x = (3+-sqrt(9+4*10))/2 = 3/2+-7/2# (by the quadratic formula)
Normally we would be done, however originally we had #log(x-3)# in the equation, meaning we have the restriction #x>3#, as the logarithm function is undefined in the reals for #x<=0#.
As #3/2-7/2 < 3# it is not a solution.
As #3/2+7/2 = 5 > 3#, it is a solution.
Thus we have the solution #x = 5#
