# How do you simplify Ln(-e) - ln(-1/e)?

Mar 23, 2016

This expression is not valid because $\ln$ function is only defined for positive numbers.

#### Explanation:

This expression is not valid because $\ln$ is not defined for negative numbers but if it was:

$\ln \left(e\right) - \ln \left(\frac{1}{e}\right)$

then you could use a formula: $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$ which would lead to:

$\ln \left(e\right) - \ln \left(\frac{1}{e}\right) = \ln \left(\frac{e}{\frac{1}{e}}\right) = \ln \left(e \cdot e\right) = \ln \left({e}^{2}\right) = 2 \cdot \ln e = 2$

In the last transformation I used an identity: $\ln \left({a}^{b}\right) = b \cdot \ln a$

Aug 10, 2017

$\ln \left(- e\right) - \ln \left(- \frac{1}{e}\right) = 2$

#### Explanation:

In order to simplify this, we need to use the difference law:

$\log a - \log b \equiv \log \left(\frac{a}{b}\right)$

$\ln \left(- e\right) - \ln \left(- \frac{1}{e}\right) = \ln \left(\frac{- e}{- \frac{1}{e}}\right) = \ln \left({e}^{2}\right) = 2 \ln e = 2$

We could also use the exponent and addition laws:

$b \log a \equiv \log \left({a}^{b}\right)$ and $\log a + \log b \equiv \log \left(a b\right)$

$\ln \left(- e\right) - \ln \left(- \frac{1}{e}\right) = \ln \left(- e\right) + \ln \left({\left(- \frac{1}{e}\right)}^{-} 1\right)$
$= \ln \left(- e\right) + \ln \left(- e\right) = \ln \left({e}^{2}\right) = 2$

Note that taking the $\log$ of negative numbers is only valid over the complex field.

Aug 16, 2017

If we are working in $\mathbb{C}$ and using principal values of $\ln \left(z\right)$, then it simplifies to $2$

#### Explanation:

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

$\ln \left(- \frac{1}{e}\right) = - 1 + \pi i$

$\ln \left(- e\right) - \ln \left(- \frac{1}{e}\right) = \left(1 + \pi i\right) - \left(- 1 + \pi i\right) = 2 + 0 i = 2$