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

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: ${\log}_{8} \left(\frac{1}{64}\right) = - 2$ and ${\log}_{8} \left({\left(\frac{1}{64}\right)}^{- 1}\right) = 2$.
By definition, for $x > 0$, ${\log}_{8} \left(x\right)$ is the unique real number satisfying the condition that ${8}^{{\log}_{8} \left(x\right)} = x$. Since ${8}^{y} > 0$ for all real $y$, the quantity $\log \left(- \frac{1}{64}\right)$ has no real number answer.
For the other examples given, since ${8}^{- 2} = \frac{1}{{8}^{2}} = \frac{1}{64}$ and ${8}^{2} = 64$, it follows that ${\log}_{8} \left(\frac{1}{64}\right) = - 2$ and ${\log}_{8} \left({\left(\frac{1}{64}\right)}^{- 1}\right) = 2$.