# How do you expand ln (e^3/(xy))?

Jul 16, 2016

$\ln \left({e}^{3} / \left(x y\right)\right) = - \ln x - \ln y + 3$

#### Explanation:

To expand, we can use the properties of logarithms.

The properties of logarithms are:

$\ln \left(u \cdot v\right) = \ln \left(u\right) + \ln \left(v\right)$

$\ln \left(\frac{u}{v}\right) = \ln \left(u\right) - \ln \left(v\right)$

$\ln \left({u}^{n}\right) = n \cdot \ln \left(u\right)$

Also remember that $\ln \left(e\right) = 1$.

Rewriting our expression yields

$\ln \left({e}^{3} / \left(x y\right)\right) = \ln \left({e}^{3}\right) - \ln \left(x y\right)$

$= 3 \cancel{\ln \left(e\right)} - \left[\ln x + \ln y\right]$

$= 3 - \ln x - \ln y$

$= - \ln x - \ln y + 3$