How do you evaluate #log_8 (-1/64)#?

1 Answer
Bill K.
Aug 4, 2016

As stated, there is no answer since there is no real number to which 8 can be raised to give a negative number. The following equations are true, however, if you are interested: #log_{8}(1/64)=-2# and #log_{8}((1/64)^{-1})=2#.

Explanation:

By definition, for #x>0#, #log_{8}(x)# is the unique real number satisfying the condition that #8^{log_{8}(x)}=x#. Since #8^{y}>0# for all real #y#, the quantity #log(-1/64)# has no real number answer.

For the other examples given, since #8^{-2}=1/(8^2)=1/64# and #8^{2}=64#, it follows that #log_{8}(1/64)=-2# and #log_{8}((1/64)^{-1})=2#.

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