How do you solve Log_2 x + log_4 x + log_8 x = 11 log2x+log4x+log8x=11?

1 Answer

x=64x=64

Explanation:

As log_ab=logb/logalogab=logbloga, we can write log_2x+log_4x+log_8x=11log2x+log4x+log8x=11 as

logx/log2+logx/log4+logx/log8=11logxlog2+logxlog4+logxlog8=11

or logx/log2+logx/(2log2)+logx/(3log2)=11logxlog2+logx2log2+logx3log2=11

or logx/log2(1+1/2+1/3)=11logxlog2(1+12+13)=11

or logx/log2((6+3+2)/6)=11logxlog2(6+3+26)=11

or logx/log2 (11/6)=11logxlog2(116)=11

or logx/log2=11xx6/11=6logxlog2=11×611=6

i.e. logx=6log2=log2^6logx=6log2=log26

and x=2^6=64x=26=64