How do you expand $ln (\frac{e^{3}}{x y})$ $ln (e^3/(xy))$ ?

Question

1975 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-16T23:30:06

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

To expand, we can use the properties of logarithms.

The properties of logarithms are:

$ln (u \cdot v) = ln (u) + ln (v)$ $ln(u*v) = ln(u) + ln(v)$

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

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

Also remember that $ln (e) = 1$ $ln(e) = 1$ .

Rewriting our expression yields

$ln (\frac{e^{3}}{x y}) = ln (e^{3}) - ln (x y)$ $ln(e^3/(xy)) = ln(e^3) - ln(xy)$

$= 3cancel(ln(e)) - [ln x + ln y]$

$=3 - ln x - ln y$

$= -ln x - ln y + 3$

How do you expand ln (e^3/(xy))ln(e3xy)ln (e^3/(xy))?