已知2(log以1/2为底x)²+7log以1/2为底x+3≤0,求函数y=(log以2为底x/2)
问题描述:
已知2(log以1/2为底x)²+7log以1/2为底x+3≤0,求函数y=(log以2为底x/2)
乘以(log以1/2为底4/x)的最大值和最小值,
答
∵2(log以1/2为底x)²+7log以1/2为底x+3≤0∴-3≤log1/2 x≤-1/2∴1/2≤log2 x≤3∴-0.5≤log2 x -1.5≤1.50≤(log2 x -1.5)²≤9/4∵y=(log以2为底x/2)乘以(log以1/2为底4/x)=(log2 x-1)(log2 x-2)=(log2 ...