y=[log(2)(x/8)][log(2)(2x)] 在[1/2,4]的值域如题2都是底数 x/8 2x都是真数

[log(4)(2x)][log(2x)(3y)][log(3y)(z)]=log(8x)[(8x)]^2,则z=? 已知log（2）[log（3)(log(4) x)] = log(3) [log(4) (log(2) y)]=0,求x+y的值 可以说log a x +log a y=log a 2log(m)y-log(m)x-log(m)z log（x+5）=log（x）+log（2） log（8）+log（x）=log（24） 2log（x）=log(2x)+log(3) 若log底数2[log底数3(log底数4)]若log底数2[log底数3(log底数4（x）)]=log底数3[log底数4(log底数2(y))]=log底数4[log底数2(log底数3z)]=0,则x+y+z= log 9 X/Y=1/2(log 3 X - log 3 Y) log 9 X/Y=1/4 3log 3 X =2log 3 Y find X and Y log(2x)+log(3y)-log(2z) 怎么化简?log3(x^2-2x-6)=2 log(3x+6)=1+log(x)求解,求每步详...log(2x)+log(3y)-log(2z) 怎么化简?log3(x^2-2x-6)=2 log(3x+6)=1+log(x)求解, y=[log(2)(x/8)][log(2)(2x)] x在[1/2,4]的值域如题 怎么画出y=log|x+2|和y=log|x|的图像 log₂log₃log₄ X=log₃log₄log₂ Y=log₄log₂log₃ Z=0X+Y+Z=?为什么是71呀? 函数y=log^2 x,x∈(0,8]的值域是(log以2为底) 函数y=log^2(2)(x)+log(2)x^2+2的值域是多少? 换底公式的使用!为什么 y=log(1/2) X = - log(2)X 函数y=log(2^2)x-log（2）x的单调递减区间 函数y=log(1/2)2x+log(1/3)x的单调区间为 log(4)y+log(2y)=9怎么解 用C语言计算log(5)log(3)-log(2))/sin(π/3)=#include #includeint main(){int a=5,b=3,c=2;double y;y=(log(a)log(b)-log(c))/sin(π/b);printf((log(5)log(3)-log(2))/sin(π/3)=%ld,y);return 0;}x这哪里错了?