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 \(b + n.\)
+129845 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
