# How do you find the exact value of log_3 (-9)?

Dec 28, 2016

$= 2 + i \frac{k}{\ln} 3 \pi , k = 1 , \pm 3 , \pm 5. . . , \pm \left(2 n + 1\right) + \ldots .$

#### Explanation:

${\log}_{b} x$ is complex of unlimited values, if x < 0.

Here,

${\log}_{3} \left(- 9\right)$

$= {\log}_{3} \left(\left({\left(\pm i\right)}^{2}\right) \left({3}^{2}\right)\right)$, where $i = \sqrt{- 1}$

$= 2 \left({\log}_{3} \left(\pm\right) i + {\log}_{3} 3\right)$

$= 2 \left(\ln \frac{\pm i}{\ln} 3 + 1\right)$, using ${\log}_{b} a = {\log}_{c} \frac{a}{\log} _ c b$

=2(ln(e^(ik/2pi)/ln3 +1), k = +-1, +--3, +- 5+ ...+- (2n+1),+....

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

using $\ln {e}^{z} = z$

Dec 28, 2016

${\log}_{3} \left(- 9\right)$ is not defined within Real numbers.
As a Complex value I think the answer is
$\textcolor{w h i t e}{\text{XXX}} {\log}_{3} \left(- 9\right) = 2 + \frac{\pi}{\ln} \left(3\right) i$

#### Explanation:

For Real numbers the $\log$ function is only defined when its argument is non-negative.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For Complex Numbers:

It has been a long time since I've done this kind of thing so could someone check my results.

One of the common results of Euler's Formula: ${e}^{i x} = \cos \left(x\right) + i \cdot \sin \left(x\right)$
is that
$\textcolor{w h i t e}{\text{XXX}} {e}^{i \pi} = - 1$
from which it follows that ${\log}_{e} \left(- 1\right) = i \pi$

Also we will need to remember the "change of base" formula for logarithms:
$\textcolor{w h i t e}{\text{XXX}} {\log}_{a} \left(p\right) = \frac{{\log}_{b} \left(p\right)}{{\log}_{b} \left(a\right)}$

....and here we go!

${\log}_{3} \left(- 9\right) = {\log}_{3} \left(9 \times \left(- 1\right)\right)$
$\textcolor{w h i t e}{\text{XXX}} = {\log}_{3} \left(9\right) + {\log}_{3} \left(- 1\right)$

$\textcolor{w h i t e}{\text{XXX}} = 2 + {\log}_{3} \left(- 1\right)$

color(white)("XXX")=2+ (log_e (-1))/(log_e(3)

$\textcolor{w h i t e}{\text{XXX}} = 2 + \frac{i \pi}{{\log}_{e} \left(3\right)}$

or using the common notation of $\ln \left(x\right)$ for ${\log}_{e} \left(x\right)$
and separating the $i$ factor from the second term:
$\textcolor{w h i t e}{\text{XXX}} = 2 + \frac{\pi}{\ln} \left(3\right) i$