log2^x=0X为多少

log2^x=0X为多少 若log2(log3(log2 x)等于0那么x的负二分之一次方为多少? log2（2^x+1)log2(2^(x+1)+2)=2的解为多少 log2^(x^log2^x)=? log2(-x)=x+ 函数f(x)=1/log2(-x^2 4x-3)的定义域为多少 log4(log2^x)=log2(log4^x)求log2^x log2(x-1)+log2(x+2)反函数为g(x) g(2)=?log2(x-1)+log2(x+2)反函数为g(x) g(2)=? [log2 1]+[log2 2]+[log2 3]+[log2 4]+[log2 5]+...+[log2 1024]=?[x]表示不超过x的最大整数2为底 答案是8204 log2 (x+1)+log2 x=log6 log2^x-1=0,求x 求f(x)=log2(3^x+1)的值域为多少 log2[log2/1(log2x)]=0 求 x log2(x)=-1 log2 (x + 3) + log2(x + 2) = 1log2 (x + 3) + log2(x + 2) = 1 设f(log2(x))=2^x(x>0),则f(3)的值为___. 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为底(x+1)