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

It is

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

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Index from and log form can be used interchangeably.

Consider it an exponential equation with the same bases.

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To evaluate ( \log_{64} \left(\frac{1}{8}\right) ), use the property of logarithms which states that ( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} ), where ( \log_c ) denotes the logarithm base ( c ).

In this case, we can rewrite ( \frac{1}{8} ) as ( 64^{-1} ) since ( \frac{1}{8} = 64^{-1} ). Therefore, ( \log_{64} \left(\frac{1}{8}\right) = \frac{\log_{64}(64^{-1})}{\log_{64}(64)} ).

Since ( \log_{64}(64) = 1 ) (since (64^1 = 64)) and ( \log_{64}(64^{-1}) = -1 ) (since (64^{-1} = \frac{1}{64}) and (64^{-1}) corresponds to a logarithm with base 64), we have:

[ \log_{64} \left(\frac{1}{8}\right) = \frac{-1}{1} = -1 ]

Therefore, ( \log_{64} \left(\frac{1}{8}\right) = -1 ).

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