# How do you evaluate log_14 (-1)?

Aug 9, 2016

$\frac{\left(4 n + 1\right) \pi}{\ln} 14 i , n = 0 , \pm 1 , \pm 2. \pm 3 , \ldots$

#### Explanation:

The values are complex.

${\log}_{14} \left(- 1\right)$

$= \ln \frac{- 1}{\ln} 14$

$= \ln \frac{{i}^{2}}{\ln} 14$

$= 2 \ln \frac{i}{\ln} 14$

$= 2 \ln {e}^{i \left(2 n \pi + \frac{\pi}{2}\right)} / \ln 14 , n = 0 , \pm 1. \pm 2 , \pm 3 , . .$.

$= \frac{\left(4 n + 1\right) \pi i}{\ln} 14 , n = 0 , \pm 1. \pm 2 , \pm 3 , . .$

Aug 9, 2016

${\log}_{14} \left(- 1\right) = i \frac{\pi \pm 2 k \pi}{\log} _ e 14$

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

#### Explanation:

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

${14}^{z} = - 1$

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

${14}^{x} {14}^{i y} = - 1$ so we have

${14}^{x} = 1 \to x = 0$ and
${14}^{i y} = - 1$

We know

$14 = {e}^{{\log}_{e} 14}$ so

${14}^{i y} = {e}^{i y {\log}_{e} 14} = \cos \left(y {\log}_{e} 14\right) + i \sin \left(y {\log}_{e} 14\right) = - 1$.

(We used de Moivre's identity ${e}^{i \phi} = \cos \left(\phi\right) + i \sin \left(\phi\right)$)

This condition is attained for

$y {\log}_{e} 14 = \pi \pm 2 k \pi$ with $k = 0 , 1 , 2 , 3 , 4 , \cdots$

so

$y = \frac{\pi \pm 2 k \pi}{\log} _ e 14$

Finally

${\log}_{14} \left(- 1\right) = i \frac{\pi \pm 2 k \pi}{\log} _ e 14$