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,，试求αβ的值

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

Answer 2

设log3(x)=t,
f(x)=(㏒3 (x/27))[㏒3(3x)]
=(log3(x)- log3(27))( log3(3) +log3(x))
=( log3(x)-3)(1+ log3(x))
=(t-3)(1+t)=t²-2t-3，
方程f(x)+m=0有两根α,β,即方程t²-2t-3+m=0有两根log3(α), log3(β),
根据韦达定理，两根之和log3(α)+ log3(β)=2，
即log3(αβ)=2，αβ=9.