Question

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

Answer 1

`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`