# 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

Aug 24, 2016

$- \frac{1}{2}$

#### Explanation:

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.

${\log}_{a} b = c \text{ " harr " } {a}^{c} = b$

${\log}_{64} \left(\frac{1}{8}\right) = x \text{ " harr " } {64}^{x} = \frac{1}{8}$

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

${64}^{x} = \frac{1}{8}$

${\left({8}^{2}\right)}^{x} = {8}^{-} 1$

${8}^{2 x} = {8}^{-} 1$

$2 x = - 1$

$x = - \frac{1}{2}$

In the format ${\log}_{64} \left(\frac{1}{8}\right)$, the question being asked is
"Which power of 64 gives $\left(\frac{1}{8}\right)$?

${64}^{- \frac{1}{2}} = \frac{1}{8}$