Question Detail

Let A = log(x+1) and B = log(y - 2log3). Then equations: A + 2B = log8, A - B = log6. Subtract the second from the first: 3B = log8 - log6 = log(8/6) = log(4/3). So B = (1/3)log(4/3) = log((4/3)^(1/3)). Then A = log6 + B = log6 + log((4/3)^(1/3)) = log(6 * (4/3)^(1/3)). This is messy. Testing options: If x=3, y=2: log(x+1)=log4, y-2log3 = 2 - 2log3, not a nice number. Option D is x=3, y=2. If x=2, y=3: log3, y-2log3=3-2log3. Not simplifying to give log8 and log6. Likely the intended simpler system had no "2log3" inside. If the equations were log(x+1)+2log y = log8 and log(x+1)-log y = log6, then adding gives 2log(x+1)=log48 => x+1=√48=4√3, not integer. Solving: from second, log(x+1)=log6+log y, substitute: log6+log y+2log y=log8 => log6+3log y=log8 => log y³=log(8/6)=log(4/3) => y³=4/3, not integer. So options are likely for a different system. Given the options, D (x=3,y=2) is common in such problems.