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

Question

Ariana Alvarez · Accepted Answer

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

Madelyn Anderson · Answer

#-1/2#

There is a relationship between 8 and 64 in that #8^2 = 64# But how does it apply in this log format.

Index from and log form can be used interchangeably.

#log_a b = c " " harr " " a^c = b#

#log_64 (1/8) = x " " harr " " 64^x = 1/8#

Consider it an exponential equation with the same bases.

#64^x = 1/8#

#(8^2)^x = 8^-1#

#8^(2x) = 8^-1#

#2x = -1#

#x = -1/2#

In the format #log_64 (1/8)#, the question being asked is "Which power of 64 gives #(1/8)#?

#64^(-1/2) = 1/8#

Greyson Brown · Answer

undefined 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 $.