Question

x-y^2-(x-y)×7-8

已知x,y属于R,且y=x^2,求证log2(2^x+2^y）>7/8~~如题~~
标准答案上写log2(2^x+2^y)=log22^（根号下x+y)=log22^[(x+y+2)/2]
谁能解释下为什么~~
谢谢了~~~
在线等

Answer 1

2^x+2^y
=2^x+2^(x^2)
>=2*(2^x*2^(x^2))^(1/2)
=2^[(x^2+x+2)/2]
所以
㏒2 (2x+2y)>=
log2{2^[(x^2+x+2)/2]}=(x^2+x+2)/2>=7/8