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How do you evaluate #log_64 (1/2)#?

1 Answer

Sep 12, 2016

#-1/6#

Explanation:

Recall that #log_2(8)=3, => 2^3=8#

let #log_64(1/2)=x#

#64^x=1/2 #

#64=2^6#
#64^x=2^(6x)#
#1/2=2^-1#

#2^(6x)=2^-1#
#6x=-1#
#x=-1/6#

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