急求解！！！！！对数方程

7 = log(2)x + 2log(4)x + log(8)x
= log(2)x + 2[log(2)x]/2 + log(2)x/3
= (7/3)log(2)x,

log(2)x = 3
x = 2^3 = 8

log2x+2log4x+log8x=7
log2x+2*log2x/(log2)4+log2x/(log2)8=7其中log后面的数字是底数。。。）
log2x+log2x+log2x/3=7
7log2x=7
log2x=1
x=2

将log的底数化出来。如log8=1/3log2 如此化简得
log2x+1/2*2log2x+1/3log2x=7
7/3log2x=7
log2x=3
x=2^3
x=8