# Question #8be2a

Feb 12, 2017

$\ln \left(a b\right) - 1$

#### Explanation:

Consider the sum of the terms, $\ln \left(\frac{1}{e}\right) + \ln \left(a b\right)$

The function $\ln$ here is defined as a function that for whatever value you enter, you get the supposed number of times you're supposed to multiply the number $e$ with itself.

In simple terms, $\ln \left(x\right)$ equals $y$, where $y$ is such a number that when $e \cdot e \cdot e \cdot \ldots . . \cdot e \cdot e$ for $y$ times (or in short ${e}^{y}$) gives back $x$ to us.

Take note of the first term given in the sum. $\ln \left(\frac{1}{e}\right)$.
$\frac{1}{e}$ can be re-written as ${e}^{-} 1$
So, $\ln \left(\frac{1}{e}\right) = \ln \left({e}^{-} 1\right)$

Another important identity of $\ln$ function is that if we have a $y = {x}^{m}$, then $\ln \left(y\right) = \ln \left({x}^{m}\right) = m \ln x$ $\forall m \in \mathbb{R}$

So, $\ln \left({e}^{-} 1\right) = - 1 \cdot \ln e$ and if you remember what I typed up in the third paragraph, then you'll realize that $\ln e = 1$, so $\ln \left(\frac{1}{e}\right) = - 1$

$\ln \left(a b\right)$ can't be simplified further than it is, so we'll have to keep it that way.

Replacing the term for $\ln \left(\frac{1}{e}\right)$ with what we've got, we come to the conclusion as written in the "answers" part.