SOLUTION: The answer to the inequality log(x2-7x) < log(3-x) + log(2 is (A) -1 < x < 0 (B) x > 7 or x < 0 (C) x > 7) or -1 < x < 6 (D) -1 < x < 6 (E) -1 < x < 3

First, the domain for this inequality is the set of real numbers x such that

    x^2 -7x > 0  and  3-x > 0, 

or    { x < 0 & x < 3 }  U  {x > 7 & x < 3 }.


Of the last two sets under the union sign, the second set  {x > 7 & x < 3 }  is empty; 
so, the domain is the set { x < 0 }.   (1)


Next, the given inequality is equivalent to 

    log(x^2 -7x) < log (2(3-x))


which implies (due to the monotonicy of the logarithm function)

    x^2 - 7x < 6 - 2x.


What follows, is the solution procedure for this inequality.

    x^2 - 5x - 6 < 0,

    (x-6)*(x+1) < 0.


The last inequality has the solution set  { -1 < x < 6 }.    (2)


To get the final answer, we shoud take the intersection of the set (2) with the domain set (1).


The intersection is the set { -1 < x < 0 },  or,  in the interval form,  (-1,0).


ANSWER.  The solution to given inequality is the set { -1 < x < 0 },  or,  in the interval form,  (-1,0).