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How do you solve #Log_2 x + log_4 x + log_8 x = 11 #?

1 Answer

Answer:

#x=64#

Explanation:

As #log_ab=logb/loga#, we can write #log_2x+log_4x+log_8x=11# as

#logx/log2+logx/log4+logx/log8=11#

or #logx/log2+logx/(2log2)+logx/(3log2)=11#

or #logx/log2(1+1/2+1/3)=11#

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

or #logx/log2 (11/6)=11#

or #logx/log2=11xx6/11=6#

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

and #x=2^6=64#