请问如何证明log(2)(x)+x < 4^(x-1),(x>=2)
请问如何证明log(2)(x)+x < 4^(x-1),(x>=2)
1/log(2)x+1/log(3)x+1/log(4)x=1,x=?
(高一)若x满足2(log(1/2)x)^2-14log(4)x+3≤0,求f(x)=[log(2)(x/2)]*{lo
log(0.5x)(2)-log(0.5x^3)(x^2)=log(0.5x^3)(4)
[( log(x^2) + log(x^4) )/ log(100 x)] = 4 求X
已知log(2)[log(3)(log(4) x)] = log(3) [log(4) (log(2) y)]=0,求x
log(X+5)+log(X+2)=1
log(x+5)=log(x)+log(2) log(8)+log(x)=log(24) 2log(x)=log(2x)
一、已知f(x)=log底下2(x^2+2x+5),g(x)=log底下1/2(x^2-3x+2)
解方程 log2(x+14)+log(x+2)=log(x+6)+3
log 8下标 (X-6)+ log 8小标 (x+6)=2 求X
请画出下列图像:log 1/2 (-x) y=log 1/3 /x/