高一一道对数最值问题已知log以
已知log(9)x = [log(3)y]^2，(1) 如果x = 3y，求X、Y的值； (2) x、y为何值时，X/Y取得最小值，此最小值是多少？
log(9)x = [log(3)y]^2
log(3)x/log(3)9 = [log(3)y]^2
(1/2)log(3)x = [log(3)y]^2 …… (*)
(1/2)log(3)3y = [log(3)y]^2
(1/2)log(3)3 + (1/2)log(3)y = [log(3)y]^2
设 s = log(3)y，原式变为
s^2 - (1/2)s - 1/2 = 0
(s - 1)(s + 1/2) = 0
log(...全部
已知log(9)x = [log(3)y]^2，(1) 如果x = 3y，求X、Y的值； (2) x、y为何值时，X/Y取得最小值，此最小值是多少？
log(9)x = [log(3)y]^2
log(3)x/log(3)9 = [log(3)y]^2
(1/2)log(3)x = [log(3)y]^2 …… (*)
(1/2)log(3)3y = [log(3)y]^2
(1/2)log(3)3 + (1/2)log(3)y = [log(3)y]^2
设 s = log(3)y，原式变为
s^2 - (1/2)s - 1/2 = 0
(s - 1)(s + 1/2) = 0
log(3)y1 = s1 = 1，y = 3，log(3)y2 = s2 = -1/2，y2 = √3/3
所以x1 = 3y1 = 9，x2 = 3y2 = √3
log(3)(x/y) = log(3)x - log(3)y = 2[log(3)y]^2 - log(3)y … (引用*式结果)
设 t = log(3)y，则原式 = 2t^2 - t
= 2[t^2 - (1/2)t]
= 2[t^2 - (1/2)t + (-1/4)^2 - (-1/4)^2]
= 2[(t - 1/4)^2 - 1/16]
可见，当t = 1/4时，log(3)(x/y)有最小值，且最小值是:-1/8。
。收起
