解2log2^(x-5)=log2^(x-1)+1
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解2log2^(x-5)=log2^(x-1)+1解2log2^(x-5)=log2^(x-1)+1解2log2^(x-5)=log2^(x-1)+12log2^(x-5)=log2^(x-1)+1l
解2log2^(x-5)=log2^(x-1)+1
解2log2^(x-5)=log2^(x-1)+1
解2log2^(x-5)=log2^(x-1)+1
2log2^(x-5)=log2^(x-1)+1
log2^(x-5)^2=log2^[2*(x-1)]
(x-5)^2=2(x-1)
x^2-12x+27=0
(x-3)(x-9)=0
x=3,x=9
代入检验
x=3使得log2^(x-5)无意义
所以x=9
x=9
x-5＞0,x-1＞0得x＞5
2log<2>(x-5)=log<2>(x-1)+1
log<2>(x-5)²=log<2>(x-1)+log<2>2
log<2>(x²-10x+25)=log<2>(2x-2)
x²-10x+25=2x-2
x²-12x+27=0
(x-3)(x-9)=0
x=9
原方程的解为x=9
解2log2^(x-5)=log2^(x-1)+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)
解方程log2(2-x)=log2(x-1)+1
log2^(x^log2^x)=?
解方程log2(x+14)-log2(x+6)=3-log2(x+2)
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2log2^(9^x-5)=log2^(3^x-2)+2
log2(9^x-5)=log2(3^x-2)+2怎么解?用图像可以吗
解-log2^[9^(x-1)-5]=-log2^[3^(x-1)-2]-2
解方程log2(9^x - 5)=log2(3^x - 2)的平方 详细!
解-log2^[9^(x-1)-5]=-log2^[3^(x-1)]-2
log2 4+log2 (x+2)怎么算?log2 4+log2 (x+2)=log2 (4x+8)
log2(x-2)=log4(5-x)
解方程：log2(9^x - 5)=log2(3^x - 2)+2解方程log2(9^x - 5)=log2(3^x - 2)+2希望有详细过程.注：log2中的2为log的下标
log2(2x)*log2(x)=2 x=?
f(x)=log2(x/8)*log2(2/x)