Question 134904
Remember that {{{log(b,x)+log(b,y)=log(b,(x*y))}}} and that since {{{log(b,x)=y}}}=>{{{b^y=x}}}, {{{log(b,b)=1}}}.


So:  {{{log(4,12)=log(4,(3*4))=log(4,3)+log(4,4)=log(4,3)+1}}}


That and remembering the fact that {{{log(b,x)-log(b,y)=log(b,(x/y))}}} should get you through the rest of this one.