How do you evaluate #log_64 (1/8)#?

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Answer 1 · 2016-08-23T01:23:57

It is

#log_64 (1/8)=log_64 1-log_64 8=-log8/log64=- log8/(log8^2)=-1/2#

Answer 2 · 2016-08-24T22:37:05

#-1/2#

There is a relationship between 8 and 64 in that #8^2 = 64#
But how does it apply in this log format.

Log form and index from are interchangeable.

#log_a b = c " " harr " " a^c = b#

#log_64 (1/8) = x " " harr " " 64^x = 1/8#

Treat it as an exponential equation and make the bases the same.

#64^x = 1/8#

#(8^2)^x = 8^-1#

#8^(2x) = 8^-1#

#2x = -1#

#x = -1/2#

In the format #log_64 (1/8)#, the question being asked is
"Which power of 64 gives #(1/8)#?

#64^(-1/2) = 1/8#