How do you evaluate #log_64 2#?

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Answer 1 · 2018-04-06T07:07:44

#log_64 2=1/6#

Let #log_64 2=t#

then #64^t=2#

or #(2^6)^t=2^1#

or #2^(6t)=2^1#

i.e. #6t=1#

and #t=1/6#

Hence #log_64 2=1/6#

Answer 2 · 2018-04-06T07:11:44

#log_64 2 =1/6#

#log_64 2=x#

#64^x=2#

#64^(1/2)=8# (square root)

#8^(1/3)=2# (cube root)

#64^((1/2)*(1/3))=2#

#64^(1/6)=2#

#log_64 2 =1/6#