Question

lg(7*2^x+8)>=log(√10) 2^x,求函数f(x)=log(1/2)x*log(1/2)x/4的最小值

Answer 1

lg(7*2^x+8)≥log(√10) 2^x,
即lg(7*2^x+8)≥log(10) [2^(2x)],
7*2^x+8≥2^(2x)
2^(2x)- 7*2^x-8≤0，-1≤2^x≤8，x≤3.
则log(1/2)x≥log(1/2)3，
设log(1/2)x=t, t≥log(1/2)3.
f(x)=log(1/2)x*log[(1/2)x/4]
= log(1/2)x*[ log(1/2)x- log(1/2)4]
= log(1/2)x*[ log(1/2)x+2]
= t^2+2t=(t+1)^2-1
注意到t≥log(1/2)3.
log(1/2)4＜log(1/2)3＜log(1/2)2，即-2＜log(1/2)3＜-1，
所以t=-1(即x=2)时，函数取到最小值-1.