設log3(x)=t
x∈[1/27,1/9]
則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
f(t)+m=0
t^2-2t-3+m=0
由根與系數(shù)的關系
t1+t2=2
log3(b)+log3(d)=2
log3(bd)=2
bd=9