# How do you evaluate log_4 (-1/64)?

Aug 9, 2016

${\log}_{4} \left(- \frac{1}{64}\right) = - 3 + i \frac{\pi \pm 2 k \pi}{\log} _ e 4$

for $k = 0 , 1 , 2 , \cdots$

#### Explanation:

Let us investigate the complex solutions. We know by the logarithm definition

${4}^{z} = - \frac{1}{64} = - {4}^{- 3}$

Supposing now $z = x + i y$ we have

${4}^{x} {4}^{i y} = - {4}^{- 3}$ so we have

${4}^{x} = {4}^{- 3} \to x = - 3$ and
${4}^{i y} = - 1$

We know

$4 = {e}^{{\log}_{e} 4}$ so

${4}^{i y} = {e}^{i y {\log}_{e} 4} = \cos \left(y {\log}_{e} 4\right) + i \sin \left(y {\log}_{e} 4\right) = - 1$. This condition is attained for

$y {\log}_{e} 4 = \pi \pm 2 k \pi$ with $k = 0 , 1 , 2 , 3 , 4 , \cdots$

so

$y = \frac{\pi \pm 2 k \pi}{\log} _ e 4$

Finally

${\log}_{4} \left(- \frac{1}{64}\right) = - 3 + i \frac{\pi \pm 2 k \pi}{\log} _ e 4$