Question

已知x属于[1/27,1/9],函数f(x)=(㏒3 x/27)[㏒3(3x)].若方程f(x)+m=0有两根α,β,试

Answer 1

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

f(t)+m=0

t^2-2t-3+m=0

由根与系数的关系

t1+t2=2

log3(b)+log3(a)=2

log3(ba)=2

ba=9