y=log(4)(1-2x+x^2) =log(2)[(1-x)^2] /log(2)(4) =2[log(2)|1-x|] /2 这几步不太理解,注：题中是以4为底和以2为底

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 已知log（2）[log（3)(log(4) x)] = log(3) [log(4) (log(2) y)]=0,求x+y的值 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, [log(4)(2x)][log(2x)(3y)][log(3y)(z)]=log(8x)[(8x)]^2,则z=? y=log(4,x^(1/2)) + log(4,2) 的反函数 y=[log(2)(x/8)][log(2)(2x)] x在[1/2,4]的值域如题 换底公式的使用!为什么 y=log(1/2) X = - log(2)X 函数y=log(1/2)2x+log(1/3)x的单调区间为 log 3 (4+x+4y)^2=0 联立 log 2 (x+1)^2=1+3log 2 y^2 求 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 1/log（2）x+1/log（3）x+1/log（4）x=1,x=? 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= 求函数y=log（2）4x*log（2）2x在(1/4 log(X+5)+log(X+2)=1 log(2)(x)log(2)(4x)-3log(2)(x)-12=0 (log(2)(x))^3-4(log(4)(x))^2+log(16)(x)=0 y=log(4)(1-2x+x^2) =log(2)[(1-x)^2] /log(2)(4) =2[log(2)|1-x|] /2 这几步不太理解,注：题中是以4为底和以2为底