作业帮log2^x*log2^2x=2 这个要怎么算?、
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log2^x*log2^2x=2 这个要怎么算?、
log2^x*log2^2x=2 这个要怎么算?、
log2^x*log2^2x=2 这个要怎么算?、
令a=log2^x
则log2^2x=log2^2+log2^x=1+a
所以a(a+1)=2
a²+a-2=0
a=-2,a=1
log2^x=-2,log2^x=1
所以x=1/4,x=2
log2 (x + 3) + log2(x + 2) = 1log2 (x + 3) + log2(x + 2) = 1
log2^x*log2^2x=2 这个要怎么算?、
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)
log2(2x)*log2(x)=2 x=?
f(x)=log2(x/8)*log2(2/x)
|2x-log2^x|
|[log2(x)]^2-3log2(x)+1|
化简：(log2(x/4))*(log2(x/2))
求导 y=log2(x^2)-log2(x)
求解log2(3x)=log2(2x+1)
解方程log2(2-x)=log2(x-1)+1
解2log2^(x-5)=log2^(x-1)+1
log2^(x^log2^x)=?
解方程log2(x+14)-log2(x+6)=3-log2(x+2)
log2(2X-1)
log2(x^2)图像