对数的运算问题-log a 1/x=-(log a 1-log a x)=log a x
对数的运算问题-log a 1/x=-(log a 1-log a x)=log a x
对数函数的问题.log a x 与log (a^-1) x的关系.
x=log a 2
设a>0,a≠1,x,y满足log以a为底x的对数+3log以x为底a的对数-log以x为底y的对数=3.
设log(a)(x+y)=根号三,log(a)x=1,求log(a)y
[log (底数为a) x]=[1\2log (底数为a) b ] - [ log (底数为a)c]求X
log a(1/2次方) x等于2log a
log以a为底(x-4)的对数>log以a为底(2x-1)对数
已知函数f(x)=log以a为底(1-x)的对数+log以a为底(x+3)的对数(0
f(x)=|log(a)(x)-1|+|2log(a)(x)|,求使f(x)<2的x范围,
证明对数运算法则(1)log(a)(MN)=log(a)(M)+log(a)(N); (2)log(a)(M/N)=
2log(b)x=log(a)x+log(c)x