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The sum \(\log_4(11)+\log_8(11)+\log_{64}(11)\) can be expressed in the form \(\log_b(n),\) where \(b\) and \(n\) are positive integers, and \(b\) is as small as possible. Find \(b + n.\)

 Apr 10, 2019
 #1
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log4(11) + log8(11) + log64(11)

 

Note that we can use the change-of-base theorem to write

 

log11/ log4  + log 11/ log 8   +  log 11 / 64  =

 

log 11/ log2^2    + log 11 / log 2^3  +  log 11 / log  2^6   =

 

log 11/ [2 log 2]  +  log 11 / [ 3 log 2 ] + log 11 /  6 log(2)  =

 

 

[ 1/2 + 1/3 + 1/6]  log 11 / log 2   =

 

[ 3/6 + 2/6 + 1/6 ]  log 11 / log 2  =

 

[ 1 ]   log 11 / log 2  =

 

Write back as

 

log 2 11

 

So  b + n  =   13

 

 

cool cool cool

 Apr 10, 2019

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