设log3(x)=t
x∈[1/27,1/9]
则t∈[-3,-2]
f(x)=log3(x/27)*log3(3x)
=[log3(x)-log3(27)]*[log3(x)+log3(3)]
=(log3(x)-3)*(log3(x)+1)
即f(t)=(t-3)*(t+1)=t^2-2t-3=(t-1)^2-4
对称轴t=1,所以,当t=-3时,有最大值=12,当t=-2时有最小值=5
f(t)+m=0
t^2-2t-3+m=0
由根与系数的关系
t1+t2=2
即log3(a)+log3(b)=2
log3(ab)=2
ab=9