# Can you take the natural log of a negative number?

Jan 25, 2016

If you are only working in real numbers, no, you cannot take the natural log of a negative number. If you are working in complex (imaginary) numbers, however, it is possible. Explanation below.

#### Explanation:

Notice that ${e}^{i \pi} = - 1$ so in some sense you might like to define:

$\ln \left(- 1\right) = \pi i$.

Combining this with $\ln \left(a b\right) = \ln \left(a\right) + \ln \left(b\right)$ you might then define:

$\ln \left(- x\right) = \ln \left(x\right) + \pi i$ when $x > 0$

However, notice that with this definition:

$\ln \left(- 1\right) + \ln \left(- 1\right) = \ln \left(1\right) + \pi i + \ln \left(1\right) + \pi i = 2 \pi i$

Whereas:

$\ln \left(- 1 \cdot - 1\right) = \ln \left(1\right) = 0$

So the $\ln \left(a b\right) = \ln \left(a\right) + \ln \left(b\right)$ rule does not quite work for Complex logarithms.

So what's going on?

The problem is that while ${e}^{x} : \mathbb{R} \to \left(0 , \infty\right)$ is a one-one function as a Real valued function, the function ${e}^{z} : \mathbb{C} \to \mathbb{C} \text{\} \left\{0\right\}$ is many to one.

So ${e}^{z}$ does not have a well defined inverse $\ln \left(z\right) : \mathbb{C} \text{\} \left\{0\right\} \to \mathbb{C}$ unless we restrict the domain of ${e}^{z}$ or the range of $\ln \left(z\right)$.

We can define a principal Complex logarithm consistent with the most common definition of $A r g \left(z\right) \in \left(- \pi , \pi\right]$ as follows:

$\ln \left(r \left(\cos \theta + i \sin \theta\right)\right) = \ln \left(r\right) + i \theta$

where $r > 0$ and $\theta \in \left(- \pi , \pi\right]$

This principal Complex logarithm satisfies:

${e}^{\ln \left(z\right)} = z$ for any $z \in \mathbb{C} \text{\} \left\{0\right\}$

$\ln \left({e}^{z}\right) = z + 2 \pi k i$ for some integer $k$ for any $z \in \mathbb{C}$

Feb 2, 2016

If you are restricted to Real numbers, then "no".
If you are working with Complex numbers see George C.'s response.

#### Explanation:

[I've included this answer only because of uncertainty of the mathematics level at which the question was asked.]

If you could take the (natural) log of a negative number (say $m < 0$)
then
$\textcolor{w h i t e}{\text{XXX}} \ln m = k$
would mean
$\textcolor{w h i t e}{\text{XXX}} {e}^{k} = m < 0$

Since $e \approx 2.7 > 0$
$\textcolor{w h i t e}{\text{XXX}} {e}^{k} > 0$ for all Real values of $k$