Question

a>1且a^(lgb)=2^(1/4)求log2(ab)的最小值

Answer 1

a^(lgb)=2^(1/4)

两边取常用对数,得

lga*lgb=1/4lg2

所以由“正数的算术平均数大于几何平均数”,

(lga+lgb)/2>=(lga*lgb)^(1/2)=1/2*(lg2)^1/2,

从而log2(ab)=(lga+lgb)/lg2>=1/(lg2)^1/2,

log2(ab)的最小值是1/(lg2)^1/2