解log2(x^2-5x+8)=1
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解log2(x^2-5x+8)=1解log2(x^2-5x+8)=1解log2(x^2-5x+8)=1log2(x^2-5x+8)=1x²-5x+8=2x²-5x+6=0(x-2)
解log2(x^2-5x+8)=1
解log2(x^2-5x+8)=1
解log2(x^2-5x+8)=1
log2(x^2-5x+8)=1
x²-5x+8=2
x²-5x+6=0
(x-2)(x-3)=0
x=2或x=3
解2log2^(x-5)=log2^(x-1)+1
解log2(x^2-5x+8)=1
解不等式log2(4x+8)>log2(2x+1)
解方程log2(2-x)=log2(x-1)+1
解方程：log2(x+4)+log2(x-1)=1+log2(x+8)
log2 (x + 3) + log2(x + 2) = 1log2 (x + 3) + log2(x + 2) = 1
[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
解-log2^[9^(x-1)-5]=-log2^[3^(x-1)-2]-2
解-log2^[9^(x-1)-5]=-log2^[3^(x-1)]-2
f(x)=log2(x/8)*log2(2/x)
log2(x^2-5x+8)=1
log2(x – 2 ) + 2 = log2(5)log2(x – 2 ) + 2 = log2(5)
log2(x – 2 ) + 2 = log2(5)log2(x – 2 ) + 2 = log2(5)
log2 4+log2 (x+2)怎么算?log2 4+log2 (x+2)=log2 (4x+8)
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
解方程：log2(x)+log2(x-8)=7
解不等式：|log2（x-1）|+log2（5-x）
log2(x+1)+log2(x+8)=3的解集是?