log的数学题2 + log 2(为底)4= 3x-1
log的数学题2 + log 2(为底)4= 3x-1
log以2为底的(log以3为底的(log以4为底的x))=log以3为底的(log以4为底的(log以2为底的y))=
已知log(2)[log(3)(log(4) x)] = log(3) [log(4) (log(2) y)]=0,求x
1/log(2)x+1/log(3)x+1/log(4)x=1,x=?
log(9)x+log(x^2)3=1的解为?
log底数7[log底数3(log底数2X)]=0,那么x^1/2的值是
对数计算1.{log(4)(3)+ log(8) (3)} X {log(3)(2)+log(9)(2)}2.{log(
y=(log以3为底x)的方-(2log以3为底的x)+2 此中x∈[1/3,9],求值域
函数y=log以a为底x,当log以a为底(x^2-x+1)小于等于log以a为底3/4成立时,a的取值范围.
log(0.5x)(2)-log(0.5x^3)(x^2)=log(0.5x^3)(4)
log以5为底(3x-2)>log以5为底(x+1)
计算:(log以4为底3的对数+log以8为底3的对数)(log以3为底2的对数+log以9为底2的对数)- log以½