How do you write #8^(-2/3)=1/4# in logarithmic form?

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Answer 1 · 2018-06-17T08:35:30

# log_8(1/4)=-2/3#

definition of a logarithm

#log_ab=c=>a^c=b#

we have

#8^(-2/3)=1/4#

comparing this with the definition

#a=8, b=1/4, c=-2/3#

#:. log_8(1/4)=-2/3#