log2(x^-x-2)>2 怎么算
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log2(x^-x-2)>2怎么算log2(x^-x-2)>2怎么算log2(x^-x-2)>2怎么算请问log后的2是它的底数吗?如果是的话,log2(x^-x-2)>2log2(x^-x-2)>l
log2(x^-x-2)>2 怎么算
log2(x^-x-2)>2 怎么算
log2(x^-x-2)>2 怎么算
请问log后的2是它的底数吗?如果是的话,
log2(x^-x-2)>2
log2(x^-x-2)>log2 4
∵log2 x在(0,正无穷)上单调递增,
x^-x-2这个应该怎样理解啊?
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已知函数f(x)=log2^ ( x/4 ) ×log2^ (2x) (1)解不等式f(x)>0;(2)当x∈【1,4】时,求f(x)的值域f(x)=log2(2x)×log2(x/4)=[(log2 2)+(log2 x)] ×[(log2 x) -(log2 4)]=[1+(log2 x)] ×[(log2 x) -2]=(log2 x)² - (log2 x) -2
log2(2x)*log2(x)=2 x=?
f(x)=log2(x/8)*log2(2/x)