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

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Answer 1 · 2016-08-23T01:23:57

It is

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

Answer 2 · 2016-08-24T22:37:05

$-1/2$

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

Log form and index from are interchangeable.

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

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

Treat it as an exponential equation and make the bases the same.

$64^x = 1/8$

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

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

$2x = -1$

$x = -1/2$

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

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

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