已知logax+3logxa-logxy=3(大于1),若y的最小值是8求a和x

问题描述:

已知logax+3logxa-logxy=3(大于1),若y的最小值是8求a和x
1个回答 分类:数学 2014-11-09

问题解答:

我来补答
logax+3logxa-logxy=3
logax=3/logax-3=logay/logax
IF logax=t logay=g(x)
g(x)=t^2-3t+3
a>1 y min=8 => g(x)min
g(x)=ax^2+bx+c a>0
x=-b/2a y=4ac-b^2/4a
=> x=3/2 y=3/4
loga8=g(x)=3/4=>a^(3/4)=8=>a^(3/4*4/3)=8^(4/3)=>a=2^4
logax=t=3/2=>log2^4 x=3/2=>2^[4*(3/2)]=x=>x=2^6
a=2^4
x=2^6
Do you understand this problem?
Please give me your opinion.
Thanks,
Yoga
 
 
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