4的logx3份之1减1等于2份之1 求x是多少?
4的log3份之1减1等于2份之1 求x是多少? 4^[log(1/3)-1]=1/2 ===> 4^[log(1/3)]/4=1/2 ===> 4^[log(1/3)]=4*(1/2)=2 ===> (2^2)^[log(1/3)]=2 ===> 2^[2log(1/3)]=2^1 ===> 2log(1/3)=1 ===> log(1/3)=(1/2) ===> x^(1/2)=1/3 ===> x=(1/3)^2=1/9
4^(log1/3-1)=1/2 :4^(log1/3)/4=1/2 4^log1/3=2 2^(2log1/3)=2 2log1/3=1 log1/3=1/2 x^1/2=1/3 x=(1/3) x=1/9
问:怎样求函数y等于log<2份之1>(负x平方加2x加8)的值域 要解过程,
答:怎样求函数y等于log<2份之1>(负x平方加2x加8)的值域 y=log<1/2>(-x^2+2x+8) 首先确定定义域,-x^2+2x+8>0 ===> x...详情>>
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